vWith the Calculus as a key, Mathematics can be successfully applied to the
explanation of the course of Nature – WHITEHEAD v
13.1 Introduction
This chapter is an introduction to Calculus. Calculus is that
branch of mathematics which mainly deals with the study
of change in the value of a function as the points in the
domain change. First, we give an intuitive idea of derivative
(without actually defining it). Then we give a naive definition
of limit and study some algebra of limits. Then we come
back to a definition of derivative and study some algebra
of derivatives. We also obtain derivatives of certain
standard functions.
13.2 Intuitive Idea of Derivatives
Physical experiments have confirmed that the body dropped
from a tall cliff covers a distance of 4.9t
2
metres in t seconds,
i.e., distance s in metres covered by the body as a function of time t in seconds is given
by s = 4.9t
2
.
The adjoining Table 13.1 gives the distance travelled in metres at various intervals
of time in seconds of a body dropped from a tall cliff.
The objective is to find the veloctiy of the body at time t = 2 seconds from this
data. One way to approach this problem is to find the average velocity for various
intervals of time ending at t = 2 seconds and hope that these throw some light on the
velocity at t = 2 seconds.
Average velocity between t = t
1
and t = t
2
equals distance travelled between
t = t
1
and t = t
2
seconds divided by (t
2
– t
1
). Hence the average velocity in the first
two seconds
13Chapter
LIMITS AND DERIVATIVES
Sir Issac Newton
(1642-1727)
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282 MATHEMATICS
=
2 1
2 1
Distance travelled between 2 0
Time interval ( )
t and t
t t
= =
−
=
(
)
( )
19.6 0
9.8 /
2 0
m
m s
s
−
=
−
.
Similarly, the average velocity between t = 1
and t = 2 is
(
)
( )
19.6 – 4.9
2 1
m
s
−
= 14.7 m/s
Likewise we compute the average velocitiy
between t = t
1
and t = 2 for various t
1
. The following
Table 13.2 gives the average velocity (v), t = t
1
seconds and t = 2 seconds.
Table 13.2
t
1
0 1 1.5 1.8 1.9 1.95 1.99
v 9.8 14.7 17.15 18.62 19.11 19.355 19.551
From Table 13.2, we observe that the average velocity is gradually increasing.
As we make the time intervals ending at t = 2 smaller, we see that we get a better idea
of the velocity at t = 2. Hoping that nothing really dramatic happens between 1.99
seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just
above 19.551m/s.
This conclusion is somewhat strengthened by the following set of computation.
Compute the average velocities for various time intervals starting at t = 2 seconds. As
before the average velocity v between t = 2 seconds and t = t
2
seconds is
=
2
2
Distance travelled between 2 seconds and
seconds
2
t
t −
=
2
2
Distance travelled in seconds Distance
travelled in 2 seconds
2
t
t
−
−
t s
0
0
1 4.9
1.5 11.025
1.8 15.876
1.9 17.689
1.95 18.63225
2 19.6
2.05 20.59225
2.1 21.609
2.2 23.716
2.5 30.625
3 44.1
4 78.4
Table 13.1
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LIMITS AND DERIVATIVES 283
=
2
2
Distance travelled in seconds 19.6
2
t
t
−
−
The following Table 13.3 gives the average velocity v in metres per second
between t = 2 seconds and t
2
seconds.
Table 13.3
t
2
4 3 2.5 2.2 2.1 2.05
2.01
v 29.4 24.5 22.05 20.58 20.09 19.845 19.649
Here again we note that if we take smaller time intervals starting at t = 2, we get
better idea of the velocity at t = 2.
In the first set of computations, what we have done is to find average velocities
in increasing time intervals ending at t = 2 and then hope that nothing dramatic happens
just before t = 2. In the second set of computations, we have found the average velocities
decreasing in time intervals ending at t = 2 and then hope that nothing dramatic happens
just after t = 2. Purely on the physical grounds, both these sequences of average
velocities must approach a common limit. We can safely conclude that the velocity of
the body at t = 2 is between 19.551m/s and 19.649 m/s. Technically, we say that the
instantaneous velocity at t = 2 is between 19.551 m/s and 19.649 m/s. As is
well-known, velocity is the rate of change of displacement. Hence what we have
accomplished is the following. From the given data of distance covered at various time
instants we have estimated the rate of
change of the distance at a given instant
of time. We say that the derivative of
the distance function s = 4.9t
2
at t = 2
is between 19.551 and 19.649.
An alternate way of viewing this
limiting process is shown in Fig 13.1.
This is a plot of distance s of the body
from the top of the cliff versus the time
t elapsed. In the limit as the sequence
of time intervals h
1
, h
2
, ..., approaches
zero, the sequence of average velocities
approaches the same limit as does the
sequence of ratios
Fig 13.1
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284 MATHEMATICS
3 3
1 1 2 2
1 2 3
C BC B C B
, ,
AC AC AC
, ...
where C
1
B
1
= s
1
– s
0
is the distance travelled by the body in the time interval h
1
= AC
1
,
etc. From the Fig 13.1 it is safe to conclude that this latter sequence approaches the
slope of the tangent to the curve at point A. In other words, the instantaneous velocity
v(t) of a body at time t = 2 is equal to the slope of the tangent of the curve s = 4.9t
2
at
t = 2.
13.3 Limits
The above discussion clearly points towards the fact that we need to understand limiting
process in greater clarity. We study a few illustrative examples to gain some familiarity
with the concept of limits.
Consider the function f(x) = x
2
. Observe that as x takes values very close to 0,
the value of f(x) also moves towards 0 (See Fig 2.10 Chapter 2). We say
(
)
0
lim 0
x
f x
→
=
(to be read as limit of f (x) as x tends to zero equals zero). The limit of f (x) as x tends
to zero is to be thought of as the value f (x) should assume at x = 0.
In general as x → a, f (x) → l, then l is called limit of the function f (x) which is
symbolically written as
(
)
lim
x a
f x l
→
=
.
Consider the following function g(x) = |x|, x
≠
0. Observe that g(0) is not defined.
Computing the value of g(x) for values of x very
near to 0, we see that the value of g(x) moves
towards 0. So,
0
lim
x
→
g(x) = 0. This is intuitively
clear from the graph of y = |x| for x
≠
0.
(See Fig 2.13, Chapter 2).
Consider the following function.
( )
2
4
, 2
2
x
h x x
x
−
= ≠
−
.
Compute the value of h(x) for values of
x very near to 2 (but not at 2). Convince yourself
that all these values are near to 4. This is
somewhat strengthened by considering the graph
of the function y = h(x) given here (Fig 13.2).
Fig 13.2
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LIMITS AND DERIVATIVES 285
In all these illustrations the value which the function should assume at a given
point x = a did not really depend on how is x tending to a. Note that there are essentially
two ways x could approach a number a either from left or from right, i.e., all the
values of x near a could be less than a or could be greater than a. This naturally leads
to two limits – the right hand limit and the left hand limit. Right hand limit
of a
function f(x) is that value of f(x) which is dictated by the values of f(x) when x tends
to a from the right. Similarly, the left hand limit. To illustrate this, consider the function
( )
1, 0
2, 0
x
f x
x
≤
=
>
Graph of this function is shown in the Fig 13.3. It is
clear that the value of f at 0 dictated by values of f(x) with
x ≤ 0 equals 1, i.e., the left hand limit of f (x) at 0 is
0
lim ( ) 1
x
f x
→
=
.
Similarly, the value of f at 0 dictated by values of
f (x) with x > 0 equals 2, i.e., the right hand limit of f (x)
at 0 is
0
lim ( ) 2
x
f x
+
→
=
.
In this case the right and left hand limits are different, and hence we say that the
limit of f (x) as x tends to zero does not exist (even though the function is defined at 0).
Summary
We say
lim
x a
→
–
f(x) is the expected value of f at x = a given the values of f near
x to the left of a. This value is called the left hand limit of f at a.
We say
lim ( )
x a
f x
+
→
is the expected value of f at x = a given the values of
f near x to the right of a. This value is called the right hand limit of f(x) at a.
If the right and left hand limits coincide, we call that common value as the limit
of f(x) at x = a and denote it by
lim
x a
→
f(x).
Illustration 1 Consider the function f(x) = x + 10. We want to find the limit of this
function at x = 5. Let us compute the value of the function f(x) for x very near to 5.
Some of the points near and to the left of 5 are 4.9, 4.95, 4.99, 4.995. . ., etc. Values
of the function at these points are tabulated below. Similarly, the real number 5.001,
Fig 13.3
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286 MATHEMATICS
5.01, 5.1 are also points near and to the right of 5. Values of the function at these points
are also given in the Table 13.4.
Table 13.4
From the Table 13.4, we deduce that value of f(x)
at x = 5 should be greater than
14.995 and less than 15.001 assuming nothing dramatic happens between x = 4.995
and 5.001. It is reasonable to assume that the value of the f(x) at x = 5 as dictated by
the numbers to the left of 5 is 15, i.e.,
(
)
–
5
lim 15
x
f x
→
=
.
Similarly, when x approaches 5 from the right, f(x) should be taking value 15, i.e.,
(
)
5
lim 15
x
f x
+
→
=
.
Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are
both equal to 15. Thus,
(
)
(
)
(
)
5
5 5
lim lim lim 15
x
x x
f x f x f x
− +
→
→ →
= = =
.
This conclusion about the limit being equal to 15 is somewhat strengthened by
seeing the graph of this function which is given in Fig 2.16, Chapter 2. In this figure, we
note that as x approaches 5 from either right or left, the graph of the function
f(x) = x +10 approaches the point (5, 15).
We observe that the value of the function at x = 5 also happens to be equal to 15.
Illustration 2 Consider the function f(x) = x
3
. Let us try to find the limit of this
function at x = 1. Proceeding as in the previous case, we tabulate the value of f(x) at
x near 1. This is given in the Table 13.5.
Table 13.5
From this table, we deduce that value of f(x) at x = 1 should be greater than
0.997002999 and less than 1.003003001 assuming nothing dramatic happens between
x 0.9 0.99 0.999 1.001 1.01 1.1
f(x) 0.729 0.970299 0.997002999 1.003003001 1.030301 1.331
x 4.9 4.95 4.99 4.995 5.001 5.01 5.1
f(x) 14.9 14.95 14.99 14.995 15.001 15.01 15.1
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LIMITS AND DERIVATIVES 287
x = 0.999 and 1.001. It is reasonable to assume that the value of the f(x) at x = 1 as
dictated by the numbers to the left of 1 is 1, i.e.,
(
)
1
lim 1
x
f x
−
→
=
.
Similarly, when x approaches 1 from the right, f(x) should be taking value 1, i.e.,
(
)
1
lim 1
x
f x
+
→
=
.
Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are
both equal to 1. Thus,
(
)
(
)
(
)
1
1 1
lim lim lim 1
x
x x
f x f x f x
− +
→
→ →
= = =
.
This conclusion about the limit being equal to 1 is somewhat strengthened by
seeing the graph of this function which is given in Fig 2.11, Chapter 2. In this figure, we
note that as x approaches 1 from either right or left, the graph of the function
f(x) = x
3
approaches the point (1, 1).
We observe, again, that the value of the function at x = 1 also happens to be
equal to 1.
Illustration 3 Consider the function f(x) = 3x. Let us try to find the limit of this
function at x = 2. The following Table 13.6 is now self-explanatory.
Table 13.6
x 1.9 1.95 1.99 1.999
2.001 2.01 2.1
f(x) 5.7 5.85 5.97 5.997 6.003 6.03 6.3
As before we observe that as x approaches 2
from either left or right, the value of f(x) seem to
approach 6. We record this as
(
)
(
)
(
)
2
2 2
lim lim lim 6
x
x x
f x f x f x
− +
→
→ →
= = =
Its graph shown in Fig 13.4 strengthens this
fact.
Here again we note that the value of the function
at x = 2 coincides with the limit at x = 2.
Illustration 4 Consider the constant function
f(x) = 3. Let us try to find its limit at x = 2. This
function being the constant function takes the same
Fig 13.4
2020-21
288 MATHEMATICS
value (3, in this case) everywhere, i.e., its value at points close to 2 is 3. Hence
(
)
(
)
(
)
2 2
2
lim lim lim 3
x x
x
f x f x f x
+
→ →
→
= = =
Graph of f(x) = 3 is anyway the line parallel to x-axis passing through (0, 3) and
is shown in Fig 2.9, Chapter 2. From this also it is clear that the required limit is 3. In
fact, it is easily observed that
(
)
lim 3
x a
f x
→
=
for any real number a.
Illustration 5 Consider the function f(x) = x
2
+ x. We want to find
(
)
1
lim
x
f x
→
. We
tabulate the values of f(x) near x = 1 in Table 13.7.
Table 13.7
x 0.9 0.99 0.999 1.01 1.1 1.2
f(x) 1.71 1.9701 1.997001 2.0301 2.31 2.64
From this it is reasonable to deduce that
(
)
(
)
(
)
1
1 1
lim lim lim 2
x
x x
f x f x f x
− +
→
→ →
= = =
.
From the graph of f(x) = x
2
+ x
shown in the Fig 13.5, it is clear that as x
approaches 1, the graph approaches (1, 2).
Here, again we observe that the
1
lim
x
→
f (x) = f (1)
Now, convince yourself of the
following three facts:
2
1 1 1
lim 1, lim 1 and lim 1 2
x x x
x x x
→ → →
= = + =
Then
2 2
1 1 1
lim lim 1 1 2 lim
x x x
x x x x
→ → →
+ = + = = +
.
Also
( ) ( )
2
1 1 1 1
lim . lim 1 1.2 2 lim 1 lim
x x x x
x x x x x x
→ → → →
+ = = = + = +
.
Fig 13.5
2020-21
LIMITS AND DERIVATIVES 289
Illustration 6 Consider the function f(x) = sin
x.
We are interested in
2
lim sin
x
x
π
→
,
where the angle is measured in radians.
Here, we tabulate the (approximate) value of f(x) near
2
π
(Table 13.8). From
this, we may deduce that
(
)
(
)
(
)
2
2 2
lim lim lim 1
x
x x
f x f x f x
− +
π
π π
→
→ →
= = =
.
Further, this is supported by the graph of f(x) = sin x which is given in the Fig 3.8
(Chapter 3). In this case too, we observe that
2
lim
x
π
→
sin x = 1.
Table 13.8
x
0.1
2
π
−
0.01
2
π
−
0.01
2
π
+
0.1
2
π
+
f(x) 0.9950 0.9999 0.9999 0.9950
Illustration 7 Consider the function f(x) = x + cos x. We want to find the
0
lim
x
→
f (x).
Here we tabulate the (approximate) value of f(x) near 0 (Table 13.9).
Table 13.9
From the Table 13.9, we may deduce that
(
)
(
)
(
)
0
0 0
lim lim lim 1
x
x x
f x f x f x
− +
→
→ →
= = =
In this case too, we observe that
0
lim
x
→
f (x) = f (0) = 1.
Now, can you convince yourself that
[
]
0 0 0
lim cos lim lim cos
x x x
x x x x
→ → →
+ = +
is indeed true?
x – 0.1 – 0.01 – 0.001 0.001 0.01 0.1
f(x) 0.9850 0.98995 0.9989995 1.0009995 1.00995 1.0950
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290 MATHEMATICS
Illustration 8 Consider the function
( )
2
1
f x
x
=
for
0
x
>
. We want to know
0
lim
x
→
f (x).
Here, observe that the domain of the function is given to be all positive real
numbers. Hence, when we tabulate the values of f(x), it does not make sense to talk of
x approaching 0 from the left. Below we tabulate the values of the function for positive
x close to 0 (in this table n denotes any positive integer).
From the Table 13.10 given below, we see that as x tends to 0, f(x) becomes
larger and larger. What we mean here is that the value of f(x) may be made larger than
any given number.
Table 13.10
x 1 0.1 0.01 10
–n
f(x) 1 100 10000 10
2n
Mathematically, we say
(
)
0
lim
x
f x
→
= +∞
We also remark that we will not come across such limits in this course.
Illustration 9 We want to find
(
)
0
lim
x
f x
→
, where
( )
2, 0
0 , 0
2, 0
x x
f x x
x x
− <
= =
+ >
As usual we make a table of x near 0 with f(x). Observe that for negative values of x
we need to evaluate x – 2 and for positive values, we need to evaluate x + 2.
Table 13.11
From the first three entries of the Table 13.11, we deduce that the value of the
function is decreasing to –2 and hence.
(
)
0
lim 2
x
f x
−
→
= −
x – 0.1 – 0.01 – 0.001 0.001 0.01 0.1
f(x) – 2.1 – 2.01 – 2.001 2.001 2.01 2.1
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LIMITS AND DERIVATIVES 291
From the last three entires of the table we deduce that the value of the function
is increasing from 2 and hence
(
)
0
lim 2
x
f x
+
→
=
Since the left and right hand limits at 0 do not coincide,
we say that the limit of the function at 0 does not exist.
Graph of this function is given in the Fig13.6. Here,
we remark that the value of the function at x = 0 is well
defined and is, indeed, equal to 0, but the limit of the function
at x = 0 is not even defined.
Illustration 10 As a final illustration, we find
(
)
1
lim
x
f x
→
,
where
( )
2 1
0 1
x x
f x
x
+ ≠
=
=
Table 13.12
x 0.9 0.99 0.999 1.001 1.01 1.1
f(x) 2.9 2.99 2.999 3.001 3.01 3.1
As usual we tabulate the values of f(x) for x near 1. From the values of f(x) for
x less than 1, it seems that the function should take value 3 at x = 1., i.e.,
(
)
1
lim 3
x
f x
−
→
=
.
Similarly, the value of f(x) should be 3 as dic-
tated by values of f(x) at x greater than 1. i.e.
(
)
1
lim 3
x
f x
+
→
=
.
But then the left and right hand limits coincide
and hence
(
)
(
)
(
)
1
1 1
lim lim lim 3
x
x x
f x f x f x
− +
→
→ →
= = =
.
Graph of function given in Fig 13.7 strengthens
our deduction about the limit. Here, we
Fig 13.6
Fig 13.7
2020-21
292 MATHEMATICS
note that in general, at a given point the value of the function and its limit may be
different (even when both are defined).
13.3.1 Algebra of limits In the above illustrations, we have observed that the limiting
process respects addition, subtraction, multiplication and division as long as the limits
and functions under consideration are well defined. This is not a coincidence. In fact,
below we formalise these as a theorem without proof.
Theorem 1 Let f and
g be two functions such that both
lim
x a
→
f (x) and
lim
x a
→
g(x) exist.
Then
(i) Limit of sum of two functions is sum of the limits of the functions, i.e.,
lim
x a
→
[f(x) + g (x)] =
lim
x a
→
f(x) +
lim
x a
→
g(x).
(ii) Limit of difference of two functions is difference of the limits of the functions, i.e.,
lim
x a
→
[f(x) – g(x)] =
lim
x a
→
f(x) –
lim
x a
→
g(x).
(iii) Limit of product of two functions is product of the limits of the functions, i.e.,
lim
x a
→
[f(x) . g(x)] =
lim
x a
→
f(x).
lim
x a
→
g(x).
(iv) Limit of quotient of two functions is quotient of the limits of the functions (whenever
the denominator is non zero), i.e.,
( )
( )
(
)
( )
lim
lim
lim
x a
x a
x a
f x
f x
g x g x
→
→
→
=
A
Note In particular as a special case of (iii), when g is the constant function
such that g(x) =
λ
, for some real number
λ
, we have
(
)
(
)
(
)
lim . .lim
x a x a
f x f x
→ →
λ = λ
.
In the next two subsections, we illustrate how to exploit this theorem to evaluate
limits of special types of functions.
13.3.2 Limits of polynomials and rational functions A function f is said to be a
polynomial function of degree n f(x) = a
0
+ a
1
x + a
2
x
2
+. . . + a
n
x
n
, where a
i
s are real
numbers such that a
n
≠ 0 for some natural number n.
We know that
lim
x a
→
x = a. Hence
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LIMITS AND DERIVATIVES 293
(
)
2 2
lim lim . lim .lim .
x a x a x a x a
x x x x x a a a
→ → → →
= = = =
An easy exercise in induction on n tells us that
lim
n n
x a
x a
→
=
Now, let
(
)
2
0 1 2
...
n
n
f x a a x a x a x
= + + + +
be a polynomial function. Thinking
of each of
2
0 1 2
, , ,...,
n
n
a a x a x a x
as a function, we have
(
)
lim
x a
f x
→
=
2
0 1 2
lim ...
n
n
x a
a a x a x a x
→
+ + + +
=
2
0 1 2
lim lim lim ... lim
n
n
x a x a x a x a
a a x a x a x
→ → → →
+ + + +
=
2
0 1 2
lim lim ... lim
n
n
x a x a x a
a a x a x a x
→ → →
+ + + +
=
2
0 1 2
...
n
n
a a a a a a a
+ + + +
=
(
)
f a
(Make sure that you understand the justification for each step in the above!)
A function f is said to be a rational function, if f(x) =
(
)
( )
g x
h x
, where g(x) and h(x)
are polynomials such that h(x) ≠ 0. Then
( )
( )
( )
(
)
( )
( )
( )
lim
lim lim
lim
x a
x a x a
x a
g x
g x g a
f x
h x h x h a
→
→ →
→
= = =
However, if h(a) = 0, there are two scenarios – (i) when g(a) ≠ 0 and (ii) when
g(a) = 0. In the former case we say that the limit does not exist. In the latter case we
can write g(x) = (x – a)
k
g
1
(x), where k is the maximum of powers of (x – a) in g(x)
Similarly, h(x) = (x – a)
l
h
1
(x) as h (a) = 0. Now, if k > l, we have
( )
( )
( )
( ) ( )
( ) ( )
1
1
lim lim
lim
lim
lim
k
x a x a
l
x a
x a
x a
g x x a g x
f x
h x
x a h x
→ →
→
→
→
−
= =
−
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294 MATHEMATICS
=
( )
( )
( )
( )
( )
( )
1
1
1 1
lim
0.
0
lim
k l
x a
x a
x a g x
g a
h x h a
−
→
→
−
= =
If k < l, the limit is not defined.
Example 1 Find the limits: (i)
3 2
1
lim 1
x
x x
→
− +
(ii)
(
)
3
lim 1
x
x x
→
+
(iii)
2 10
1
lim 1 ...
x
x x x
→−
+ + + +
.
Solution The required limits are all limits of some polynomial functions. Hence the
limits are the values of the function at the prescribed points. We have
(i)
1
lim
x
→
[x
3
– x
2
+ 1] = 1
3
– 1
2
+ 1 = 1
(ii)
(
)
(
)
(
)
3
lim 1 3 3 1 3 4 12
x
x x
→
+ = + = =
(iii)
2 10
1
lim 1 ...
x
x x x
→−
+ + + +
( ) ( ) ( )
2 10
1 1 1 ... 1
= + − + − + + −
1 1 1... 1 1
= − + + =
.
Example 2 Find the limits:
(i)
2
1
1
lim
100
x
x
x
→
+
+
(ii)
3 2
2
2
4 4
lim
4
x
x x x
x
→
− +
−
(iii)
2
3 2
2
4
lim
4 4
x
x
x x x
→
−
− +
(iv)
3 2
2
2
2
lim
5 6
x
x x
x x
→
−
− +
(v)
2 3 2
1
2 1
lim
3 2
x
x
x x x x x
→
−
−
− − +
.
Solution All the functions under consideration are rational functions. Hence, we first
evaluate these functions at the prescribed points. If this is of the form
0
0
, we try to
rewrite the function cancelling the factors which are causing the limit to be of
the form
0
0
.
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LIMITS AND DERIVATIVES 295
(i) We have
2 2
1
1 1 1 2
lim
100 1 100 101
x
x
x
→
+ +
= =
+ +
(ii) Evaluating the function at 2, it is of the form
0
0
.
Hence
3 2
2
2
4 4
lim
4
x
x x x
x
→
− +
−
=
( )
( )( )
2
2
2
lim
2 2
x
x x
x x
→
−
+ −
=
(
)
( )
2
2
lim as 2
2
x
x x
x
x
→
−
≠
+
=
(
)
2 2 2
0
0
2 2 4
−
= =
+
.
(iii) Evaluating the function at 2, we get it of the form
0
0
.
Hence
2
3 2
2
4
lim
4 4
x
x
x x x
→
−
− +
=
(
)
(
)
( )
2
2
2 2
lim
2
x
x x
x x
→
+ −
−
=
(
)
( ) ( )
2
2
2 2 4
lim
2 2 2 2 0
x
x
x x
→
+
+
= =
− −
which is not defined.
(iv) Evaluating the function at 2, we get it of the form
0
0
.
Hence
3 2
2
2
2
lim
5 6
x
x x
x x
→
−
− +
=
(
)
( )( )
2
2
2
lim
2 3
x
x x
x x
→
−
− −
=
( )
( )
2
2
2
2
4
lim 4
3 2 3 1
x
x
x
→
= = = −
− − −
.
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296 MATHEMATICS
(v) First, we rewrite the function as a rational function.
2 3 2
2 1
3 2
x
x x x x x
−
−
− − +
=
( )
( )
2
2 1
1
3 2
x
x x
x x x
−
−
−
− +
=
( ) ( )( )
2 1
1 1 2
x
x x x x x
−
−
− − −
=
( )( )
2
4 4 1
1 2
x x
x x x
− + −
− −
=
( )( )
2
4 3
1 2
x x
x x x
− +
− −
Evaluating the function at 1, we get it of the form
0
0
.
Hence
2
2 3 2
1
2 1
lim
3 2
x
x
x x x x x
→
−
−
− − +
=
( )( )
2
1
4 3
lim
1 2
x
x x
x x x
→
− +
− −
=
(
)
(
)
( )( )
1
3 1
lim
1 2
x
x x
x x x
→
− −
− −
=
( )
1
3
lim
2
x
x
x x
→
−
−
=
( )
1 3
1 1 2
−
−
= 2.
We remark that we could cancel the term (x – 1) in the above evaluation because
1
x
≠
.
Evaluation of an important limit which will be used in the sequel is given as a
theorem below.
Theorem 2 For any positive integer n,
1
lim
n n
n
x a
x a
na
x a
−
→
−
=
−
.
Remark The expression in the above theorem for the limit is true even if n is any
rational number and a is positive.
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LIMITS AND DERIVATIVES 297
Proof Dividing (x
n
– a
n
) by (x – a),
we see that
x
n
– a
n
= (x–a) (x
n–1
+ x
n–2
a + x
n–3
a
2
+ ... +
x a
n–2
+ a
n–1
)
Thus,
lim lim
n n
x a x a
x a
x a
→ →
−
=
−
(x
n–1
+ x
n–2
a + x
n–3
a
2
+ ... + x a
n–2
+ a
n–1
)
= a
n – l
+ a a
n–2
+. . . + a
n–2
(a) +a
n–l
= a
n–1
+ a
n – 1
+...+a
n–1
+ a
n–1
(n terms)
=
1
n
na
−
Example 3 Evaluate:
(i)
15
10
1
1
lim
1
x
x
x
→
−
−
(ii)
0
1 1
lim
x
x
x
→
+ −
Solution (i) We have
15
10
1
1
lim
1
x
x
x
→
−
−
=
15 10
1
1 1
lim
1 1
x
x x
x x
→
− −
÷
− −
=
15 10
1 1
1 1
lim lim
1 1
x x
x x
x x
→ →
− −
÷
− −
= 15 (1)
14
÷ 10(1)
9
(by the theorem above)
= 15 ÷ 10
3
2
=
(ii) Put y = 1 + x, so that
1
y
→
as
0.
x
→
Then
0
1 1
lim
x
x
x
→
+ −
=
1
1
lim
–1
y
y
y
→
−
=
1 1
2 2
1
1
lim
1
y
y
y
→
−
−
=
1
1
2
1
(1)
2
−
(by the remark above) =
1
2
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298 MATHEMATICS
13.4 Limits of Trigonometric Functions
The following facts (stated as theorems) about functions in general come in handy in
calculating limits of some trigonometric functions.
Theorem 3 Let f and
g
be two real valued functions with the same domain such that
f (x) ≤ g( x) for all x in the domain of definition, For some a, if both
lim
x a
→
f(x) and
lim
x a
→
g(x) exist, then
lim
x a
→
f(x) ≤
lim
x a
→
g(x). This is illustrated in Fig 13.8.
Theorem 4 (Sandwich Theorem) Let f, g and h be real functions such that
f (x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number
a, if
lim
x a
→
f(x) = l =
lim
x a
→
h(x), then
lim
x a
→
g(x) = l. This is illustrated in Fig 13.9.
Given below is a beautiful geometric proof of the following important
inequality relating trigonometric functions.
sin
cos 1
x
x
x
< <
for
π
0
2
x
< <
(*)
Fig 13.8
Fig 13.9
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LIMITS AND DERIVATIVES 299
Proof We know that sin (–
x) = – sin
x and cos(
– x) = cos x. Hence, it is sufficient
to prove the inequality for
π
0
2
x
< <
.
In the Fig 13.10, O is the centre of the unit circle such that
the angle AOC is x radians and 0 < x <
π
2
. Line segments B A and
CD are perpendiculars to OA. Further, join AC. Then
Area of
OAC
∆
< Area of sector
OAC
< Area of
∆
OAB.
i.e.,
2
1 1
OA.CD .
π.(OA) OA.AB
2 2
π 2
x
< <
.
i.e., CD < x . OA < AB.
From
∆
OCD,
sin x =
CD
OA
(since OC = OA) and hence CD = OA sin x. Also tan x =
AB
OA
and
hence AB = OA. tan x. Thus
OA sin x < OA. x < OA. tan x.
Since length OA is positive, we have
sin x < x < tan x.
Since 0 < x <
π
2
, sinx is positive and thus by dividing throughout by sin x, we have
1<
1
sin cos
x
x x
<
. Taking reciprocals throughout, we have
sin
cos 1
x
x
x
< <
which complete the proof.
Theorem 5 The following are two important limits.
(i)
0
sin
lim 1
x
x
x
→
=
. (ii)
0
1 cos
lim 0
x
x
x
→
−
=
.
Proof (i) The inequality in (*) says that the function
sin
x
x
is sandwiched between the
function cos x and the constant function which takes value 1.
Fig 13.10
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300 MATHEMATICS
Further, since
0
lim
x
→
cos x = 1, we see that the proof of (i) of the theorem is
complete by sandwich theorem.
To prove (ii), we recall the trigonometric identity 1 – cos x = 2 sin
2
2
x
.
Then
0
1 cos
lim
x
x
x
→
−
=
2
0 0
2sin sin
2 2
lim lim .sin
2
2
x x
x x
x
x
x
→ →
=
=
0 0
sin
2
lim .limsin 1.0 0
2
2
x x
x
x
x→ →
= =
Observe that we have implicitly used the fact that
0
x
→
is equivalent to
0
2
x
→
. This
may be justified by putting y =
2
x
.
Example 4 Evaluate: (i)
0
sin 4
lim
sin 2
x
x
x
→
(ii)
0
tan
lim
x
x
x
→
Solution (i)
0
sin 4
lim
sin 2
x
x
x
→
0
sin 4 2
lim . .2
4 sin 2
x
x x
x x
→
=
=
0
sin 4 sin 2
2.lim
4 2
x
x x
x x
→
÷
=
4 0 2 0
sin 4 sin 2
2. lim lim
4 2
x x
x x
x x
→ →
÷
= 2.1.1 = 2 (as x → 0, 4x → 0 and 2x → 0)
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LIMITS AND DERIVATIVES 301
(ii) We have
0
tan
lim
x
x
x
→
=
0
sin
lim
cos
x
x
x x
→
=
0 0
sin 1
lim . lim
cos
x x
x
x x
→ →
= 1.1 = 1
A general rule that needs to be kept in mind while evaluating limits is the following.
Say, given that the limit
(
)
( )
lim
x a
f x
g x
→
exists and we want to evaluate this. First we check
the value of f (a) and g(a). If both are 0, then we see if we can get the factor which
is causing the terms to vanish, i.e., see if we can write f(x) = f
1
(x) f
2
(x) so that
f
1
(a) = 0 and f
2
(a)
≠ 0. Similarly, we write g(x) = g
1
(x) g
2
(x), where g
1
(a) = 0 and
g
2
(a) ≠ 0. Cancel out the common factors from f(x) and g(x) (if possible) and write
(
)
( )
(
)
( )
f x p x
g x q x
=
, where q(x) ≠ 0.
Then
(
)
( )
(
)
( )
lim
x a
f x p a
g x q a
→
=
.
EXERCISE 13.1
Evaluate the following limits in Exercises 1 to 22.
1.
3
lim 3
x
x
→
+
2.
π
22
lim
7
x
x
→
−
3.
2
1
lim
π
r
r
→
4.
4
4 3
lim
2
x
x
x
→
+
−
5.
10 5
1
1
lim
1
x
x x
x
→ −
+ +
−
6.
( )
5
0
1 1
lim
x
x
x
→
+ −
7.
2
2
2
3 10
lim
4
x
x x
x
→
− −
−
8.
4
2
3
81
lim
2 5 3
x
x
x x
→
−
− −
9.
0
lim
1
x
ax b
cx
→
+
+
10.
1
3
1
1
6
1
lim
1
z
z
z
→
−
−
11.
2
2
1
lim , 0
x
ax bx c
a b c
cx bx a
→
+ +
+ + ≠
+ +
12.
2
1 1
2
lim
2
x
x
x
→−
+
+
13.
0
sin
lim
x
ax
bx
→
14.
0
sin
lim , , 0
sin
x
ax
a b
bx
→
≠
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302 MATHEMATICS
15.
(
)
( )
π
sin π
lim
π π
x
x
x
→
−
−
16.
0
cos
lim
π
x
x
x
→
−
17.
0
cos 2 1
lim
cos 1
x
x
x
→
−
−
18.
0
cos
lim
sin
x
ax x x
b x
→
+
19.
0
lim sec
x
x x
→
20.
0
sin
lim , , 0
sin
x
ax bx
a b a b
ax bx
→
+
+ ≠
+
, 21.
0
lim (cosec cot )
x
x x
→
−
22.
π
2
tan 2
lim
π
2
x
x
x
→
−
23. Find
(
)
0
lim
x
f x
→
and
(
)
1
lim
x
f x
→
, where
( )
( )
2 3, 0
3 1 , 0
x x
f x
x x
+ ≤
=
+ >
24. Find
(
)
1
lim
x
f x
→
, where
( )
2
2
1, 1
1, 1
x x
f x
x x
− ≤
=
− − >
25. Evaluate
(
)
0
lim
x
f x
→
, where
( )
| |
, 0
0, 0
x
x
f x
x
x
≠
=
=
26. Find
(
)
0
lim
x
f x
→
, where
( )
, 0
| |
0, 0
x
x
x
f x
x
≠
=
=
27. Find
(
)
5
lim
x
f x
→
, where
(
)
| | 5
f x x
= −
28. Suppose
( )
, 1
4, 1
, 1
a bx x
f x x
b ax x
+ <
= =
− >
and if
1
lim
x
→
f (x) = f (1) what are possible values of a and b?
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LIMITS AND DERIVATIVES 303
29. Let a
1
, a
2
, . . .,
a
n
be fixed real numbers and define a function
(
)
(
)
(
)
(
)
1 2
...
n
f x x a x a x a
= − − −
.
What is
1
lim
x a
→
f (x) ? For some a ≠ a
1
, a
2
, ..., a
n
, compute
lim
x a
→
f (x).
30. If
( )
1, 0
0, 0
1, 0
x x
f x x
x x
+ <
= =
− >
.
For what value (s) of a does
lim
x a
→
f (x) exists?
31. If the function f(x) satisfies
(
)
2
1
2
lim
π
1
x
f x
x
→
−
=
−
, evaluate
(
)
1
lim
x
f x
→
.
32. If
( )
2
3
, 0
, 0 1
, 1
mx n x
f x nx m x
nx m x
+ <
= + ≤ ≤
+ >
. For what integers m and n does both
(
)
0
lim
x
f x
→
and
(
)
1
lim
x
f x
→
exist?
13.5 Derivatives
We have seen in the Section 13.2, that by knowing the position of a body at various
time intervals it is possible to find the rate at which the position of the body is changing.
It is of very general interest to know a certain parameter at various instants of time and
try to finding the rate at which it is changing. There are several real life situations
where such a process needs to be carried out. For instance, people maintaining a
reservoir need to know when will a reservoir overflow knowing the depth of the water
at several instances of time, Rocket Scientists need to compute the precise velocity
with which the satellite needs to be shot out from the rocket knowing the height of the
rocket at various times. Financial institutions need to predict the changes in the value of
a particular stock knowing its present value. In these, and many such cases it is desirable
to know how a particular parameter is changing with respect to some other parameter.
The heart of the matter is derivative of a function at a given point in its domain
of definition.
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304 MATHEMATICS
Definition 1 Suppose f is a real valued function and a is a point in its domain of
definition. The derivative of f at a is defined by
(
)
(
)
0
lim
h
f a h f a
h
→
+ −
provided this limit exists. Derivative of f (x) at a is denoted by f
′
(a).
Observe that f
′
(a) quantifies the change in f(x) at a with respect to x.
Example 5 Find the derivative at x = 2 of the function f(x) = 3x.
Solution We have
f
′
(2) =
(
)
(
)
0
2 2
lim
h
f h f
h
→
+ −
=
(
)
(
)
0
3 2 3 2
lim
h
h
h
→
+ −
=
0 0 0
6 3 6 3
lim lim lim3 3
h h h
h h
h h
→ → →
+ −
= = =
.
The derivative of the function 3x at x = 2 is 3.
Example 6 Find the derivative of the function f(x) = 2x
2
+ 3x – 5 at x = –1. Also prove
that f
′
(0) + 3f
′
( –1) = 0.
Solution We first find the derivatives of f(x) at x = –1 and at x = 0. We have
(
)
' 1
f
−
=
(
)
(
)
0
1 1
lim
h
f h f
h
→
− + − −
=
( ) ( ) ( ) ( )
2 2
0
2 1 3 1 5 2 1 3 1 5
lim
h
h h
h
→
− + + − + − − − + − −
=
( ) ( )
2
0 0
2
lim lim 2 1 2 0 1 1
h h
h h
h
h
→ →
−
= − = − = −
and
(
)
' 0
f
=
(
)
(
)
0
0 0
lim
h
f h f
h
→
+ −
=
( ) ( ) ( ) ( )
2 2
0
2 0 3 0 5 2 0 3 0 5
lim
h
h h
h
→
+ + + − − + −
2020-21
LIMITS AND DERIVATIVES 305
=
( ) ( )
2
0 0
2 3
lim lim 2 3 2 0 3 3
h h
h h
h
h
→ →
+
= + = + =
Clearly
(
)
(
)
' 0 3 ' 1 0
f f
+ − =
Remark At this stage note that evaluating derivative at a point involves effective use
of various rules, limits are subjected to. The following illustrates this.
Example 7 Find the derivative of sin x at x = 0.
Solution Let f(x) = sin x. Then
f ′(0) =
(
)
(
)
0
0 0
lim
h
f h f
h
→
+ −
=
(
)
(
)
0
sin 0 sin 0
lim
h
h
h
→
+ −
=
0
sin
lim 1
h
h
h
→
=
Example 8 Find the derivative of f(x) = 3 at x = 0 and at x = 3.
Solution Since the derivative measures the change in function, intuitively it is clear
that the derivative of the constant function must be zero at every point. This is indeed,
supported by the following computation.
(
)
' 0
f
=
(
)
(
)
0 0 0
0 0
3 3 0
lim lim lim 0
h h h
f h f
h h h
→ → →
+ −
−
= = =
.
Similarly
(
)
' 3
f
=
(
)
(
)
0 0
3 3
3 3
lim lim 0
h h
f h f
h h
→ →
+ −
−
= =
.
We now present a geomet-
ric interpretation of derivative of a
function at a point. Let y = f(x) be
a function and let P = (a, f(a)) and
Q = (a + h, f(a + h) be two points
close to each other on the graph
of this function. The Fig 13.11 is
now self explanatory.
Fig 13.11
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306 MATHEMATICS
We know that
( )
(
)
(
)
0
lim
h
f a h f a
f a
h
→
+ −
′
=
From the triangle PQR, it is clear that the ratio whose limit we are taking is
precisely equal to tan(QPR) which is the slope of the chord PQ. In the limiting process,
as h tends to 0, the point Q tends to P and we have
(
)
(
)
0 Q P
QR
lim lim
PR
h
f a h f a
h
→ →
+ −
=
This is equivalent to the fact that the chord PQ tends to the tangent at P of the
curve y = f(x). Thus the limit turns out to be equal to the slope of the tangent. Hence
(
)
tan
ψ
f a
′
=
.
For a given function f we can find the derivative at every point. If the derivative
exists at every point, it defines a new function called the derivative of f . Formally, we
define derivative of a function as follows.
Definition 2 Suppose f is a real valued function, the function defined by
(
)
(
)
0
lim
h
f x h f x
h
→
+ −
wherever the limit exists is defined to be the derivative of f at x and is denoted by
f
′
(x). This definition of derivative is also called the first principle of derivative.
Thus
( )
(
)
(
)
0
' lim
h
f x h f x
f x
h
→
+ −
=
Clearly the domain of definition of f
′
(x) is wherever the above limit exists. There
are different notations for derivative of a function. Sometimes f
′
(x) is denoted by
( )
(
)
d
f x
dx
or if y = f(x), it is denoted by
dy
dx
. This is referred to as derivative of f(x)
or y with respect to x. It is also denoted by D (f (x) ). Further, derivative of f at x = a
is also denoted by
( ) or
a a
d df
f x
dx dx
or even
x a
df
dx
=
.
Example 9 Find the derivative of f(x) = 10x.
Solution Since f
′
( x) =
(
)
(
)
0
lim
h
f x h f x
h
→
+ −
2020-21
LIMITS AND DERIVATIVES 307
=
(
)
(
)
0
10 10
lim
h
x h x
h
→
+ −
=
0
10
lim
h
h
h
→
=
(
)
0
lim 10 10
h→
=
.
Example 10 Find the derivative of f(x) = x
2
.
Solution We have, f ′(x) =
(
)
(
)
0
lim
h
f x h f x
h
→
+ −
=
(
)
(
)
2 2
0
lim
h
x h x
h
→
+ −
=
(
)
0
lim 2 2
h
h x x
→
+ =
Example 11 Find the derivative of the constant function f (x) = a for a fixed real
number a.
Solution We have, f ′(x) =
(
)
(
)
0
lim
h
f x h f x
h
→
+ −
=
0 0
0
lim lim 0
h h
a a
h h
→ →
−
= =
as
0
h
≠
Example 12 Find the derivative of f(x) =
1
x
Solution We have f ′(x) =
(
)
(
)
0
lim
h
f x h f x
h
→
+ −
=
0
1 1
–
( )
lim
h
x h x
h
→
+
=
(
)
( )
0
1
lim
h
x x h
h x x h
→
− +
+
=
( )
0
1
lim
h
h
h x x h
→
−
+
=
( )
0
1
lim
h
x x h
→
−
+
=
2
1
x
−
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308 MATHEMATICS
13.5.1 Algebra of derivative of functions Since the very definition of derivatives
involve limits in a rather direct fashion, we expect the rules for derivatives to follow
closely that of limits. We collect these in the following theorem.
Theorem 5 Let
f and g be two functions such that their derivatives are defined in a
common domain. Then
(i) Derivative of sum of two functions is sum of the derivatives of the
functions.
( ) ( )
( ) ( )
d d d
f x g x f x g x
dx dx dx
+ = +
.
(ii) Derivative of difference of two functions is difference of the derivatives of
the functions.
( ) ( )
( ) ( )
d d d
f x g x f x g x
dx dx dx
− = −
.
(iii) Derivative of product of two functions is given by the following product
rule.
( ) ( )
. ( ) . ( ) ( ) . ( )
d d d
f x g x f x g x f x g x
dx dx dx
= +
(iv) Derivative of quotient of two functions is given by the following quotient
rule (whenever the denominator is non–zero).
( )
2
( ) . ( ) ( ) ( )
( )
( )
( )
d d
f x g x f x g x
d f x
dx dx
dx g x
g x
−
=
The proofs of these follow essentially from the analogous theorem for limits. We
will not prove these here. As in the case of limits this theorem tells us how to compute
derivatives of special types of functions. The last two statements in the theorem may
be restated in the following fashion which aids in recalling them easily:
Let
(
)
u f x
=
and
v
= g (x). Then
( )
uv
′
=
u v uv
′ ′
+
This is referred to a Leibnitz rule for differentiating product of functions or the
product rule. Similarly, the quotient rule is
2020-21
LIMITS AND DERIVATIVES 309
u
v
′
=
2
u v uv
v
′ ′
−
Now, let us tackle derivatives of some standard functions.
It is easy to see that the derivative of the function f(x) = x is the constant
function 1. This is because
(
)
f x
′
=
(
)
(
)
0
lim
h
f x h f x
h
→
+ −
=
0
lim
h
x h x
h
→
+ −
=
0
lim1 1
h→
=
.
We use this and the above theorem to compute the derivative of
f(x) = 10x = x + .... + x (ten terms). By (i) of the above theorem
( )
df x
dx
=
d
dx
(
)
...
x x
+ +
(ten terms)
=
. . .
d d
x x
dx dx
+ +
(ten terms)
=
1 ... 1
+ +
(ten terms) = 10.
We note that this limit may be evaluated using product rule too. Write
f(x) = 10x = uv, where u is the constant function taking value 10 everywhere and
v(x) = x. Here, f(x) = 10x = uv we know that the derivative of u equals 0. Also
derivative of v(x) = x equals 1. Thus by the product rule we have
(
)
f x
′
=
( ) ( )
10 0. 10.1 10
x uv u v uv x
′ ′
′ ′
= = + = + =
On similar lines the derivative of f(x) = x
2
may be evaluated. We have
f(x) = x
2
= x .x and hence
df
dx
=
( ) ( ) (
)
. . .
d d d
x x x x x x
dx dx dx
= +
=
1. .1 2
x x x
+ =
.
More generally, we have the following theorem.
Theorem 6 Derivative of f(x) = x
n
is nx
n – 1
for any positive integer n.
Proof By definition of the derivative function, we have
( )
(
)
(
)
0
' lim
h
f x h f x
f x
h
→
+ −
=
=
(
)
0
lim
n
n
h
x h x
h
→
+ −
.
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310 MATHEMATICS
Binomial theorem tells that (x + h)
n
=
(
)
(
)
(
)
1
0 1
C C ... C
n n n n n n
n
x x h h
−
+ + +
and
hence (x + h)
n
– x
n
= h(nx
n – 1
+... + h
n – 1
). Thus
( )
df x
dx
=
( )
0
lim
n
n
h
x h x
h
→
+ −
=
(
)
1 1
0
....
lim
n n
h
h nx h
h
− −
→
+ +
=
(
)
1 1
0
lim ...
n n
h
nx h
− −
→
+ +
=
1
n
nx
−
.
Alternatively, we may also prove this by induction on n and the product rule as
follows. The result is true for n = 1, which has been proved earlier. We have
(
)
n
d
x
dx
=
(
)
1
.
n
d
x x
dx
−
=
( )
( ) ( )
1 1
. .
n n
d d
x x x x
dx dx
− −
+
(by product rule)
=
( )
(
)
1 2
1. . 1
n n
x x n x
− −
+ −
(by induction hypothesis)
=
(
)
1 1 1
1
n n n
x n x nx
− − −
+ − =
.
Remark The above theorem is true for all powers of x, i.e., n can be any real number
(but we will not prove it here).
13.5.2 Derivative of polynomials and trigonometric functions We start with the
following theorem which tells us the derivative of a polynomial function.
Theorem 7 Let f(x) =
1
1 1 0
....
n n
n n
a x a x a x a
−
−
+ + + +
be a polynomial function, where
a
i
s are all real numbers and a
n
≠
0. Then, the derivative function is given by
( )
1 2
1
( )
1 ...
n x
n n
df x
na x n a x
dx
− −
−
= + − + +
2 1
2
a x a
+
.
Proof of this theorem is just putting together part (i) of Theorem 5 and Theorem 6.
Example 13 Compute the derivative of 6x
100
– x
55
+ x.
Solution A direct application of the above theorem tells that the derivative of the
above function is
99 54
600 55 1
x x
− +
.
2020-21
LIMITS AND DERIVATIVES 311
Example 14 Find the derivative of f(x) = 1 + x + x
2
+ x
3
+... + x
50
at x = 1.
Solution
A direct application of the above Theorem 6 tells that the derivative of the
above function is 1 + 2x + 3x
2
+ . . . + 50x
49
. At x = 1 the value of this function equals
1 + 2(1) + 3(1)
2
+ .. . + 50(1)
49
= 1 + 2 + 3 + . . . + 50 =
(
)
(
)
50 51
2
= 1275.
Example 15 Find the derivative of f(x) =
1
x
x
+
Solution Clearly this function is defined everywhere except at x = 0. We use the
quotient rule with u = x + 1 and v = x. Hence u
′
= 1 and v
′
= 1. Therefore
( ) 1
df x d x d u
dx dx x dx v
+
= =
(
)
(
)
2 2 2
1 1 1
1
x x
u v uv
v x x
− +
′ ′
−
= = = −
Example 16 Compute the derivative of sin x.
Solution Let f(x) = sin x. Then
( )
df x
dx
=
(
)
(
)
(
)
(
)
0 0
sin sin
lim lim
h h
f x h f x x h x
h h
→ →
+ − + −
=
=
0
2
2cos sin
2 2
lim
h
x h h
h
→
+
(using formula for sin A – sin B)
=
0 0
sin
2
lim cos .lim cos .1 cos
2
2
h h
h
h
x x x
h→ →
+ = =
.
Example 17 Compute the derivative of tan x.
Solution Let f(x) = tan x. Then
( )
df x
dx
=
(
)
(
)
(
)
(
)
0 0
tan tan
lim lim
h h
f x h f x x h x
h h
→ →
+ − + −
=
=
(
)
( )
0
sin
1 sin
lim
cos cos
h
x h
x
h x h x
→
+
−
+
2020-21
312 MATHEMATICS
=
(
)
(
)
( )
0
sin cos cos sin
lim
cos cos
h
x h x x h x
h x h x
→
+ − +
+
=
(
)
( )
0
sin
lim
cos cos
h
x h x
h x h x
→
+ −
+
(using formula for sin (A + B))
=
( )
0 0
sin 1
lim .lim
cos cos
h h
h
h x h x
→ →
+
=
2
2
1
1. sec
cos
x
x
=
.
Example 18 Compute the derivative of f(x) = sin
2
x.
Solution We use the Leibnitz product rule to evaluate this.
(
)
( )
sin sin
df x d
x x
dx dx
=
( ) ( )
sin sin sin sin
x x x x
′ ′
= +
(
)
(
)
cos sin sin cos
x x x x
= +
2sin cos sin 2
x x x
= =
.
EXERCISE 13.2
1. Find the derivative of x
2
– 2 at x = 10.
2. Find the derivative of x at x = 1.
3. Find the derivative of 99x at x = l00.
4. Find the derivative of the following functions from first principle.
(i)
3
27
x
−
(ii)
(
)
(
)
1 2
x x
− −
(iii)
2
1
x
(iv)
1
1
x
x
+
−
5. For the function
( )
100 99 2
. . . 1
100 99 2
x x x
f x x
= + + + + +
.
2020-21
LIMITS AND DERIVATIVES 313
Prove that
(
)
(
)
1 100 0
f f
′ ′
=
.
6. Find the derivative of
1 2 2 1
. . .
n n n n n
x ax a x a x a
− − −
+ + + + +
for some fixed real
number
a.
7. For some constants a and b, find the derivative of
(i)
(
)
(
)
x a x b
− −
(ii)
(
)
2
2
ax b
+
(iii)
x a
x b
−
−
8. Find the derivative of
n n
x a
x a
−
−
for some constant a.
9. Find the derivative of
(i)
3
2
4
x
−
(ii)
(
)
(
)
3
5 3 1 1
x x x
+ − −
(iii)
(
)
3
5 3
x x
−
+
(iv)
(
)
5 9
3 6
x x
−
−
(v)
(
)
4 5
3 4
x x
− −
−
(vi)
2
2
1 3 1
x
x x
−
+ −
10. Find the derivative of cos x from first principle.
11. Find the derivative of the following functions:
(i)
sin cos
x x
(ii)
sec
x
(iii)
5sec 4cos
x x
+
(iv) cosec x (v)
3cot 5cosec
x x
+
(vi)
5sin 6 cos 7
x x
− +
(vii)
2 tan 7sec
x x
−
Miscellaneous Examples
Example 19 Find the derivative of f from the first principle, where f is given by
(i) f (x) =
2 3
2
x
x
+
−
(ii) f (x) =
1
x
x
+
Solution (i) Note that function is not defined at x = 2. But, we have
( )
( ) ( )
(
)
0 0
2 3
2 3
2 2
lim lim
h h
x h
x
f x h f x
x h x
f x
h h
→ →
+ +
+
−
+ −
+ − −
′
= =
2020-21
314 MATHEMATICS
=
(
)
(
)
(
)
(
)
( )( )
0
2 2 3 2 2 3 2
lim
2 2
h
x h x x x h
h x x h
→
+ + − − + + −
− + −
=
(
)
(
)
(
)
(
)
(
)
(
)
( )( )
0
2 3 2 2 2 2 3 2 2 3
lim
2 2
h
x x h x x x h x
h x x h
→
+ − + − − + − − +
− + −
=
( ) ( )
( )
2
0
–7 7
lim
2 2
2
h
x x h
x
→
= −
− + −
−
Again, note that the function
f
′
is also not defined at x = 2.
(ii) The function is not defined at x = 0. But, we have
(
)
f x
′
=
( ) ( )
0 0
1 1
lim lim
h h
x h x
f x h f x
x h x
h h
→ →
+ + − +
+ −
+
=
=
0
1 1 1
lim
h
h
h x h x
→
+ −
+
=
( ) ( )
0 0
1 1 1
lim lim 1
h h
x x h
h h
h x x h h x x h
→ →
− −
+ = −
+ +
=
( )
2
0
1 1
lim 1 1
h
x x h
x
→
− = −
+
Again, note that the function
f
′
is not defined at x = 0.
Example 20 Find the derivative of f(x) from the first principle, where f(x) is
(i)
sin cos
x x
+
(ii)
sin
x x
Solution (i) we have
(
)
'
f x
=
(
)
(
)
f x h f x
h
+ −
=
(
)
(
)
0
sin cos sin cos
lim
h
x h x h x x
h
→
+ + + − −
=
0
sin cos cos sin cos cos sin sin sin cos
lim
h
x h x h x h x h x x
h
→
+ + − − −
2020-21
LIMITS AND DERIVATIVES 315
=
(
)
(
)
(
)
0
sin cos sin sin cos 1 cos cos 1
lim
h
h x x x h x h
h
→
− + − + −
=
( )
(
)
0 0
cos 1
sin
lim cos sin lim sin
h h
h
h
x x x
h h
→ →
−
− +
(
)
0
cos 1
lim cos
h
h
x
h
→
−
+
=
cos sin
x x
−
(ii)
(
)
'
f x
=
(
)
(
)
(
)
(
)
0 0
sin sin
lim lim
h h
f x h f x x h x h x x
h h
→ →
+ − + + −
=
=
(
)
(
)
0
sin cos sin cos sin
lim
h
x h x h h x x x
h
→
+ + −
=
(
)
(
)
0
sin cos 1 cos sin sin cos sin cos
lim
h
x x h x x h h x h h x
h
→
− + + +
=
(
)
0
0
sin cos 1
sin
lim lim cos
h
h
x x h
h
x x
h h
→
→
−
+
(
)
0
lim sin cos sin cos
h
x h h x
→
+ +
=
cos sin
x x x
+
Example 21 Compute derivative of
(i) f(x) = sin 2x (ii) g(x) = cot x
Solution (i) Recall the trigonometric formula sin 2x = 2 sin x cos x. Thus
( )
df x
dx
=
( ) (
)
2sin cos 2 sin cos
d d
x x x x
dx dx
=
( ) ( )
2 sin cos sin cos
x x x x
′ ′
= +
(
)
(
)
2 cos cos sin sin
x x x x
= + −
(
)
2 2
2 cos sin
x x
= −
(ii) By definition, g(x) =
cos
cot
sin
x
x
x
=
. We use the quotient rule on this function
wherever it is defined.
cos
(cot )
sin
dg d d x
x
dx dx dx x
= =
2020-21
316 MATHEMATICS
=
2
(cos ) (sin ) (cos )(sin )
(sin )
x x x x
x
′ ′
−
=
2
( sin )(sin ) (cos )(cos )
(sin )
x x x x
x
− −
=
2 2
2
2
sin cos
cosec
sin
x x
x
x
+
− =−
Alternatively, this may be computed by noting that
1
cot
tan
x
x
=
. Here, we use the fact
that the derivative of tan x is sec
2
x which we saw in Example 17 and also that the
derivative of the constant function is 0.
dg
dx
=
1
(cot )
tan
d d
x
dx dx x
=
=
2
(1) (tan ) (1)(tan )
(tan )
x x
x
′ ′
−
=
2
2
(0) (tan ) (sec )
(tan )
x x
x
−
=
2
2
2
sec
cosec
tan
x
x
x
−
= −
Example 22 Find the derivative of
(i)
5
cos
sin
x x
x
−
(ii)
cos
tan
x x
x
+
Solution (i) Let
5
cos
( )
sin
x x
h x
x
−
=
. We use the quotient rule on this function wherever
it is defined.
5 5
2
( cos ) sin ( cos ) (sin )
( )
(sin )
x x x x x x
h x
x
′ ′
− − −
′
=
2020-21
LIMITS AND DERIVATIVES 317
=
4 5
2
(5 sin )sin ( cos ) cos
sin
x x x x x x
x
+ − −
=
5 4
2
cos 5 sin 1
(sin )
x x x x
x
− + +
(ii) We use quotient rule on the function
cos
tan
x x
x
+
wherever it is defined.
( )
h x
′
=
2
( cos ) tan ( cos ) (tan )
(tan )
x x x x x x
x
′ ′
+ − +
=
2
2
(1 sin ) tan ( cos )sec
(tan )
x x x x x
x
− − +
Miscellaneous Exercise on Chapter 13
1. Find the derivative of the following functions from first principle:
(i)
x
−
(ii)
1
( )
x
−
−
(iii) sin (x + 1) (iv) cos (x –
π
8
)
Find the derivative of the following functions (it is to be understood that a, b, c, d,
p, q, r and s are fixed non-zero constants and m and n are integers):
2. (x + a) 3. (px + q)
r
s
x
+
4.
( )( )
2
ax b cx d
+ +
5.
ax b
cx d
+
+
6.
1
1
1
1
x
x
+
−
7.
2
1
ax bx c
+ +
8.
2
ax b
px qx r
+
+ +
9.
2
px qx r
ax b
+ +
+
10.
4 2
cos
a b
x
x x
− +
11.
4 2
x
−
12.
( )
n
ax b
+
13.
( ) ( )
n m
ax b cx d
+ +
14. sin (x + a) 15. cosec x cot x 16.
cos
1 sin
x
x
+
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318 MATHEMATICS
17.
sin cos
sin cos
x x
x x
+
−
18.
sec 1
sec 1
x
x
−
+
19.
sin
n
x
20.
sin
cos
a b x
c d x
+
+
21.
sin( )
cos
x a
x
+
22.
4
(5sin 3cos )
x x x
−
23.
(
)
2
1 cos
x x
+
24.
(
)
( )
2
sin cos
ax x p q x
+ +
25.
(
)
(
)
cos tan
x x x x
+ −
26.
4 5sin
3 7cos
x x
x x
+
+
27.
2
cos
4
sin
x
x
π
28.
1 tan
x
x
+
29.
(
)
(
)
sec tan
x x x x
+ −
30.
sin
n
x
x
Summary
®
The expected value of the function as dictated by the points to the left of a
point defines the left hand limit of the function at that point. Similarly the right
hand limit.
®
Limit of a function at a point is the common value of the left and right hand
limits, if they coincide.
®
For a function f and a real number a,
lim
x a
→
f(x) and f (a) may not be same (In
fact, one may be defined and not the other one).
®
For functions f and g the following holds:
[
]
lim ( ) ( ) lim ( ) lim ( )
x a x a x a
f x g x f x g x
→ → →
± = ±
[
]
lim ( ). ( ) lim ( ).lim ( )
x a x a x a
f x g x f x g x
→ → →
=
lim ( )
( )
lim
( ) lim ( )
x a
x a
x a
f x
f x
g x g x
→
→
→
=
®
Following are some of the standard limits
1
lim
n n
n
x a
x a
na
x a
−
→
−
=
−
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LIMITS AND DERIVATIVES 319
0
sin
lim 1
x
x
x
→
=
0
1 cos
lim 0
x
x
x
→
−
=
®
The derivative of a function f at a is defined by
0
( ) ( )
( ) lim
h
f a h f a
f a
h
→
+ −
′
=
®
Derivative of a function f at any point x is defined by
0
( ) ( ) ( )
( ) lim
h
df x f x h f x
f x
dx h
→
+ −
′
= =
®
For functions u and v the following holds:
( )
u v u v
′ ′ ′
± = ±
( )
uv u v uv
′ ′ ′
= +
2
u u v uv
v
v
′
′ ′
−
=
provided all are defined.
®
Following are some of the standard derivatives.
1
( )
n n
d
x nx
dx
−
=
(sin ) cos
d
x x
dx
=
(cos ) sin
d
x x
dx
=−
Historical Note
In the history of mathematics two names are prominent to share the credit for
inventing calculus, Issac Newton (1642 – 1727) and G.W. Leibnitz (1646 – 1717).
Both of them independently invented calculus around the seventeenth century.
After the advent of calculus many mathematicians contributed for further
development of calculus. The rigorous concept is mainly attributed to the great
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320 MATHEMATICS
mathematicians, A.L. Cauchy, J.L.Lagrange and Karl Weierstrass. Cauchy gave
the foundation of calculus as we have now generally accepted in our textbooks.
Cauchy used D’ Alembert’s limit concept to define the derivative of a function.
Starting with definition of a limit, Cauchy gave examples such as the limit of
sin
α
α
for
α
= 0. He wrote
( ) ( )
,
y f x i f x
x i
∆ + −
=
∆
and called the limit for
0,
i
→
the “function derive’e, y′ for f ′ (x)”.
Before 1900, it was thought that calculus is quite difficult to teach. So calculus
became beyond the reach of youngsters. But just in 1900, John Perry and others
in England started propagating the view that essential ideas and methods of calculus
were simple and could be taught even in schools. F.L. Griffin, pioneered the
teaching of calculus to first year students. This was regarded as one of the most
daring act in those days.
Today not only the mathematics but many other subjects such as Physics,
Chemistry, Economics and Biological Sciences are enjoying the fruits of calculus.
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vv
v
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