LIMITS OF FUNCTIONS

1 Introduction

The limits concept is one of the most important ideas in calculus. It is the foundation of almost ofall mathematical analysis.

The limit of a function f(x) as x approaches c is the value that f(x) approaches as x approachesthe number c.

Definition 1 If f(x) get closer and closer to a number L as x gets closer and closer to c from bothsides, then L is the limit of f(x) as x approaches c. Mathematically this is written as

limx→c

f(x) = L

In limits we look at the behaviour of f(x) as x gets closer and closer to c. We are only talkingabout what happens to f(x) as x approaches c, and not about what happens when x equals c. It istherefore important to remember that limits describe the behavior of a function near to a particularpoint, not necessarily at the point. For example:

Example 1 : Consider the following function

f(x) =

{x2 + 2 when x 6= 2;

9 when x = 2

The limit of f(x) as x approaches 2 is equal to 6. Yet the function behave quite differently at x = 2as f(2) = 9.

Suppose we want to evaluate the limit

limx→0

f(x + 5)− f(5)

x, x 6= 0,

where f(x) = x2 + 3.

A straight substitution of x = 0 leads to the meaning expression

f(5)− f(5)

0=

0

0

1

and we must penetrate a little deeper. We have f(5) = 28 and f(x + 5) = (x + 5)2 + 3 =x2 + 10x + 28. Therefore,

f(x + 5)− f(5)

x=

x2 + 10x + 28− 28

x= x + 10

Therefore,

limx→0

f(x + 5)− f(5)

x= lim

x→0(x + 10) = 10

Example 2 : Find the limit of

limh→0

f(−1 + h)− f(−1)√h

, h > 0,

where f(x) = x3.

Solution We have f(−1) = −1 and f(−1 + h) = (−1 + h)3 = −1 + 3h − 3h2 + h3. So,f(−1 + h)− f(−1) = h(3− 3h + h2). Hence,

limh→0

f(−1−+h)− f(−1)√h

= limh→0

h(3− 3h + h2)√h

= limh→0

√h(3− 3h + h2) = 0

2 Limit Applications: Derivatives

Definition 2 If f is a function, the derivative of the function f , denoted by f ′ is defined by thefomular

f ′(x) = limh→0

f(x + h)− f(x)

h

In this definition, x remain fixed, while h tends to zero. If the limit does not exist for a particularvalue of x, then the functin has no derivative for that value. We also say that a function f isdiffentiable at x if the above limit exists. A function is f is simply said to be differentiable if ithas a derivative for all x in its domain.

Example 3 : Given f(x) = x2. Find the f ′(x).

Solution: We have f(x) = x2 and f(x+h) = x2+2xh+h2. Therefore, f(x+h)−f(x) = 2xh+h2.Thus,

f ′(x) = limh→0

(x + h)− f(x)

h= lim

h→0

2xh + h2

h= lim

h→0(2x + h) = 2x.

2

Example 4 : Given that f(x) = 1x, x 6= 0. Find f ′(x).

Solution: We have f(x− h)− f(x) = 1(x+h)

− 1x

= x−(x+h)x(x+h)

= −hx(x+h)

. Thus,

f ′(x) = limh→0

(x + h)− f(x)

h= lim

h→0

−h

hx(x + h)= lim

h→0

−1

(x2 + xh)=−1

x2.

3 Computation of Limits

To compute limits of complex functions we need the following theorems

Theorem 3

1. limx→c k = k, where k is any constant

2. limx→c x = c

Theorem 4 Suppose thatlimx→c

f(x) = L and limx→c

g(x) = M.

Then

1. limx→c(f(x) + g(x)) = L + M

2. limx→c(f(x)− g(x)) = L−M

3. limx→c kf(x) = kL, where k is a any constant

4. limx→c f(x)g(x) = LM

5. limx→c

(f(x)g(x)

)= L

M, provided M 6= 0

6. limx→c(f(x))n = Ln, where n is any positive integer

7. limx→cn√

f(x) = n√

L

Example 5 : Findlimx→8

(2x2 − 5x + 3)

3

Solution

limx→8

(2x2 − 5x + 3) = limx→8

(2x2 − 5x) + limx→8

3

= limx→8

(2x2)− limx→8

(5x) + limx→8

3

= 2(limx→8

(x2))− 5

(limx→8

x)

+ limx→8

3

= 2(limx→8

x)2

− 5(limx→8

)+ lim

x→83

= 2(8)2 − 5(8) + 3

= 91

Example 6 : Find

limx→−2

(x3 + 5

x2 + x− 3

)

Solution The limit of the denominator is limx→−2(x2 + x − 3) = (−2)2 − 2 − 3 = −1, which is

not zero. So, we can use rule 5 in Theorem 4. Thus,

limx→−2

(x3 + 5

x2 + x− 3

)=

lim x→ −2(x3 + 5)

lim x→ −2(x2 + x− 3)

=(−2)3 + 5

−1

=−3

−1= 3.

Example 7 : Find

limx→3

x2 − 8x + 15

x2 + 4x− 21

Solution: The limit of the denominator is

limx→−3

(x2 + 4x− 21) = (3)2 + 4(3)− 21 = 0.

So we cannot use rule 5 in Theorem 4. Instead, we if it is possible to simplify the given function.A standard method is to factorize both numerator and denomenator. We have x2 − 8x + 15 =

4

(x− 3)(x− 5) and x2 + 4x− 21 = (x− 3)(x + 7). Thus,

limx→3

x2 − 8x + 15

x2 + 4x− 21= lim

x→3

(x− 3)(x− 5)

(x− 3)(x− 5)

= limx→3

(x− 5)

(x− 5)(Cancelling the common factor x− 3)

=3− 5

3 + 7

= −1

5

Example 8 : Find

limx→5

√3x + 1− 4

x− 5

Solution The limit of the function is of the form 00. The main problem here is it might not very

clear how th find the hiden factor x− 5 of in the numerator. One techniques is to rationalize. Thatis, we use the algebraic identity

(a + b)(a− b) = a2 − b2

Thus,

limx→5

√3x + 1− 4

x− 5= lim

x→5

(√

3x + 1− 4)(√

3x + 1 + 4)

(x− 5)(√

3x + 1 + 4)

= limx→5

(√

3x + 1)2 − 16

(x− 5)(√

3x + 1 + 4)

= limx→5

3x + 1− 16

(x− 5)(√

3x + 1 + 4)

= limx→5

3x− 15

(x− 5)(√

3x + 1 + 4)

= limx→5

3(x− 5)

(x− 5)(√

3x + 1 + 4)

= limx→5

3√3x + 1 + 1

= limx→5

3(x− 5)

(x− 5)(√

3(5) + 1 + 1

= limx→5

3√3(5) + 1 + 1

=3

8

5

Example 9 : Find

limx→0

x2 sin1

x

Solution: Although the function is of the form f(x)g(x), we cannot use rule 5 in Theorem 4,because limx→0 sin 1

xdoesnot exist. To see this, as x getting smaller and smaller 1

xkeep growing

in size indefinetly!!

In this case we need the following theorem to accomplish the task.

Theorem 5 (Sandwitch Theorem) Suppose that f(x), g(x) and h(x) have property that f(x) ≤g(x) ≤ (h) for all x close to c, except possibly for x = c. If limx→c f(h) = L and limx→c h(x) =L then limx→c g(x) = L

We know−1 ≤ sin

1

x≤ 1

Thus,

−x2 ≤ x2 sin1

x≤ x2

Now limx→0−x2 = 0 and limx→0 x2 = 0. Therefore, by the Sandwitch theorem, we have

limx→0

x2 sin1

x= 0

4 One Sided Limits

Sometimes we want to study where f(x) is heading when x is heading toward a only from theright hand side of a or only from the left hand side of c; and so we have the following notions ofone-sided limits:

We writelim

x→a+f(x) = L

to mean that as x is moving toward a from the right hand side of c; the function value f(x)approaches L. We also say that the right hand limit of f as x approaches c from the right is L, andL is the right hand limit of f(x) at c.

We writelim

x→a−f(x) = L

6

to mean that as x is moving toward a from the left hand side of a, the function value f(x) ap-proaches L. We also say that the right hand limit of f as x approaches c from the left is L, and Lis called the left hand limit of f(x) at c.

It is not difficult to see that rules of limit calculations can be also applied to one-sided limits.

Example 10 : For the function

f(x) =

{1− x2 if x < 22x + 1 if x ≥ 2

Findlim

x→2−f(x) and lim

x→2+f(x)

Solution Since f(x) = 1− x2 for x < 2, we have

limx→2−

f(x) = limx→2−

(1− x2) = −3

Similarly, f(x) = 2x + 1 if x ≥ 2, so

limx→2+

f(x) = limx→2+

(2x + 1) = 5

Example 11 : For the function

f(x) =

{x(x− 3) if x < 4

x2

x+4if x ≥ 4

Findlim

x→4−f(x) and lim

x→4+f(x)

Solution Since f(x) = x(x− 3) for x < 4, we have

limx→4−

f(x) = limx→2−

(x(x− 3)) = 4

Similarly, f(x) = x2

x+4if x ≥ 4, so

limx→4+

f(x) = limx→4+

x2

x + 4=

42

8= 2

7

Theorem 6 If f is a function and c and L are numbers, then

limx→c

f(x) = L

if and only iflim

x→c−f(x) = L and lim

x→c+f(x) = L.

The theorem says that the limit of f exists of and only if both sided limits exist and they have thesame value.

Example 12 : Determine whether limx→1 f(x) exists, where

f(x) =

{x + 1 ifx < 1

−x2 + 4x− 1 if x ≥ 1

Solution Computing the one-sided limits at x = 1, we find

limx→1−

f(x) = limx→1−

(x + 1) = (1) + 1 = 2

andlim

x→1+f(x) = lim

x→1+(−x2 + 4x + 1) = −(1)2 + 4(1) + 1 = 2

Since the two one-sided limits are equal, it follows that the limit of f(x) at 1 exist and we have

limx→1−

f(x) = limx→1+

f(x) = limx→1

f(x) = 2.

5 Limits at Infinity

The symbols ∞ and −∞ are not regarded as real numbers. They are symbols to indicate that anumber increases or decreases indefinitely. When the limit of a function is ∞ or −∞ no limitexists; the symbol is used for convenience only.

Theorem 7

limx→∞

1

x= 0, lim

x→+∞

1

x= 0, lim

x→−∞

1

x= 0, and lim

x→∞

1

xn= 0 for any positive integer p

Example 13 : Evaluate

limx→∞

3x− 2

5x + 4

8

Solution:

limx→∞

3x− 2

5x + 4= lim

x→∞

3x−2x

5x+4x

= limx→∞

3xx− 2

x5xx

+ 4x

= limx→∞

3− 2x

5 + 4x

=3− 0

5 + 0

=3

5

Example 14 : Find

limx→∞

√x2 − 1

2x + 1

Solution:

limx→∞

√x2 − 1

2x + 1= lim

x→∞

√1− 1

x2

2 + 1x

=

√1− 0

2 + 0

=1

2

Example 15 : Find

limx→−∞

3x + 2

x2 − x + 1

Solution :

limx→−∞

3x + 2

x2 − x + 1= lim

x→−∞

3xx2 + 2

x2

x2

x2 − xx2 + 1

x2

= limx→−∞

3x

+ 2x2

1− 1x

+ 1x2

=0 + 0

1− 0 + 0= 0

9