Calculus: Limits & Continuity

LIMITS & CONTINUITY LIMIT OF A FUNCTION The concept of the limit of a function is the starting point of calculus. Without limits calculus does not exist. Every notion in calculus can be expressed in some forms of limits.

What is limit of a function? To understand it, let’s take the following examples : Consider the function

f x = 2x @ 3 . At x = 3, the value of the function becomes ` a

f 3 = 2.3 @ 5 = 1. Here the function is defined and real. But, how does the function ` a

behave when x gets closer to 3 but not exactly 3? Putting a few values we see that as x approaches 3, f(x) approaches 1.

sin 2x πf f f f f f f f f f f f f f f f f f f f πf f f f Consider another function f x = f we can find A If we take the interval @ f , f 3x 4 4 πf f f fπ f f f f f f f πf f f f f the value of the function at x = @ f , , etc as given below, 6 15 9 D

` a

eG πf f f f f d e sin 2 A @ 6 πf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f D E f @ = f = 0.5513, 6 πf f f f f 3 @ 6 D E πf f f f f d e sin 2 A 9 πf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f D E = 0.6138 f f = f 9 πf f f f f 3 9 F

d

f

d

E

πf f f f f f f

D

E

e sin 2 A 15 πf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f D E = 0.6473, = f 15 π 3 fffffff

15

` a sin 0f f f f f f f f f f f f which But what happens to this function at x = 0? Putting x = 0 in f, we get f 0 = f 0

is undefined. So, the function f is undefined at x = 0. But we see that it is not undefined at other x values close to 0. There it is defined and has real values. So how does the function behave there, when x goes closer to 0, but not exactly 0?

To know the behavior, we take a few values of x closer to 0 (both from x>0 and x<0) and put them in the function as given in the table below and note the behavior : 0.1

x f(x)

0.05

f(x)

0.02

0.01

0.005

0.66223 0.66555 0.66626 0.66649 0.66662 0.66665 -0.1

x

0.03

-0.05

-0.03

-0.02

-0.01

-0.005

0.66223 0.66555 0.66626 0.66649 0.66662 0.66665

2f f as x gets closer to 0 from both The value of f(x) gets closer and closer to 0.666….. or f 3 sides. ` a sin 2x f f f f f f f f f f f f f f f f 2f f , which reads “the limit of f(x) as x This behavior is written as lim f x = lim f = f xQ 0 x Q 0 3x 3

2f f f ” . It means, as the value of x gets closer and closer to 0 (i.e. x tends to 0 is f 3

2f 2f f f (or it approaches f ) . [There is approaches 0), f(x) is defined there and gets closer to f 3 3 an easier limit evaluation method to find this value, which is discussed later in trigonometric limits]. 2 xf @ 4f f f f f f f f f f f f f f f f f = 4 means as the value of x gets closer and closer to Taking another example, lim f xQ 2 x @2 2 xf @ 4f f f f f f f f f f f f f f f f f gets closer to 4 (though the function is undefined at x=2). 2, f x @2

Thus, the limit of a function f(x) at x = c gives us an idea about how this function behaves, when the value of x goes very near to c (but not equal to c). lim f x = L

xQc

` a

[read as “the limit of f(x) as x tends to c is L” ]

means

“as x approaches c, f(x) approaches L” or

“as the value of x goes very close to c, but not equal to c, f(x)goes very close to L”.

Definition of limit :

lim f x = A xQa ` a

The limit of a function

if and only if, for any chosen number ε > 0 , however small, there exists a number δ > 0 ` a such that, whenever 0 < |x @ a| < δ , then |f x @ A| < ε A Note that, the limit of a function gives an idea about how the function behaves close to a particular x value. It does not necessarily give the value of the function at x. That means, ` a

` a

lim f x and f c may or may not be equal. xQc Using the definitions we can prove the following limits : lim |x| = |c|

lim x = c

xQ c

lim a = a

xQ c

xQ c

Example : Using the definition show that lim 3x @ 1 = 5 . xQ 2

Solution :

`

a

Finding a δ : Let ε > 0 A We try to find a number δ > 0 such that, ` a whenever 0 < |x @ 2| < δ , then | 3x @ 1 @ 5| < ε A ` a Now , | 3x @ 1 @ 5| = |3x @ 6| = 3|x @ 2| To make | 3x @ 1 @ 5| < ε `

a

or

3|x @ 2| < ε we have to make |x @ 2| <

So we have to choose a δ such that 0< δ ≤ Showing that the δ works : εf f f then 3|x @ 2| < ε If 0 < |x @ 2| < 3 ` a So, | 3x @ 1 @ 5| < ε # lim 3x @ 1 = 5 xQ 2

`

a

εf f f A 3

εf f f 3

ONE SIDED LIMITS: RIGHT AND LEFT LIMITS To analyze the behavior of the function f(x) when x approaches a, we can separately look at the behavior from one side only i.e. from the left side (when xa ). lim f x = A is the left hand limit which means “as x approaches a through values less x Q a@ ` a

than a, f(x) approaches A.” Definition of Left-Hand Limit : `

a

Let f be a function defined at least on an interval c,a , then

lim f x = A

x Q a@

if and only if, for any chosen number, ε > 0 , however small, there exists a ` a number δ > 0 such that, whenever a @ δ<x
` a

lim+ f x = B is the right hand limit which means “as x approaches a through values

xQ a

` a

more than a, f(x) approaches B.” Definition of Right-Hand Limit : b

c

Let f be a function defined at least on an interval a,d , then

lim+ f x = A

xQ a

if and only if, for any chosen number, ε rel="nofollow"> 0 , however small, there exists a ` a number δ > 0 such that, whenever a < x < a + δ , then |f x @ A| < ε A

` a

The existence of limit from the left does not imply the existence of limit from the right, and vice versa.

EXISTENCE OF LIMIT AT A POINT For a function f(x), when can we say that the limit exists at a point x=a and what would be its value in relation to the left and right limits at that point? ` a

lim f x exists and its value will be equal to A, xQ a if and only if, ` a lim f x and lim+ f x both exist, and both are equal to A x Q a@ ` a

xQ a

Example : Find

Solution :

w w w w w w w

lim@ p x

xQ0

w w w w w w w

and

w w w w w w w

lim+ p x , and comment if lim p x exists. xQ 0

xQ 0

w w w w w w w

w w w w w w w lim@ p x does not exist since x<0 here and p x is not defined when x<0 A

xQ0

w w w w w w w

w w w w w w w

lim+ p x = 0 as we see that p x appproaches 0 as x approaches 0 A

xQ 0

w w w w w w w

w w w w w w w

lim p x does not exist as lim@ p x does not exist.

xQ 0

xQ0

Example : Evaluate (a) (b)

|xf @ 3| f f f f f f f f f f f f f f f f f lim@ f xQ 3 x @3 lim+

xQ3

(c)

|xf @ 3| f f f f f f f f f f f f f f f f f f x @3

|xf @ 3| f f f f f f f f f f f f f f f f f lim f x @3

xQ 3

Solution : (a) As x approaches 3 from the left, x < 3 i.e. (x-3) is negative, and |x-3| = -(x-3), hence `

a

|xf @ 3| xf @ 3f f f f f f f f f f f f f f f f f f f @ f f f f f f f f f f f f f f f f f f f f f f f f f f f lim@ = = @1 xQ 3 x @3 x @3

(b) As x approaches 3 from the right, x > 3 i.e. (x-3) is positive, and |x-3| = (x-3), hence lim+

xQ 3

|xf @ 3| @ 3f f f f f f f f f f f f f f f f f f f xf f f f f f f f f f f f f f f f = =1 x @3 x @3

|xf @ 3| |xf @ 3| f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f (c) As lim@ f ≠ lim+ , xQ 3 x @3 x @ 3 xQ 3

|xf @ 3| f f f f f f f f f f f f f f f f f f does not exist A xQ 3 x @3 lim

INFINITE LIMITS Some functions increase or decrease without bounds (that is goes towards + 1 or @ 1 ) near certain values of the independent variable x. When this happens, we say that the function has infinite limit. So we can write,

f x = @1 . The lim f x = + 1 or xlim Qa

xQa

` a

` a

unction has a vertical asymptote at x = a if either of the above limits hold true. ` a

ff xf f f f f f f f f f f f f ` a will have an infinite limit at x=a if there the limit of g(x) is A function of the form f g x zero but the limit of f(x) is non –zero. 1f f f f f f Example : Evaluate lim f xQ 0 x2

Solution :

LIMITS AT INFINITY In some cases we may need to observe the behavior of the function when the independent variable x increases or decreases without bound, that is, x Q + 1 or x Q @ 1 . If in such a case the function approaches a real number A , then we can write lim f x = A or xQ +1 ` a

lim f x = A xQ@1 ` a

and f(x) has a horizontal asymptote at y = A. 2

Example : Evaluate Solution :

3x @ 7f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f . lim f x Q + 1 5x 2 @ 2x + 3

2 3x @ 7f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim 2 1 xQ + + 3 5x @ 2x d e

=

=

=

7f f f f f f f 2 x f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f d e lim f xQ +1 2f 3f f f f f f f f f f 2 x 5@ + 2 x x d e 7f f f f f f f 3@ 2 xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f F d e lim As x approaches + 1 , xQ +1 2 3 5 @ ffff+ fff2ffff x x ` a 3f @ 0f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f a lim `

x2 3 @

xQ +1

G 1f f f f approaches 0 x

5@0 + 0

3f f = f 5

THEOREMS ON LIMITS 1 A If f x = c, where c is a constant, then xlim f x =c Qa ` a

` a

f x = A and xlim g x = B where A,B < 1 If xlim Qa Qa ` a

` a

2 A xlim kf x = kA where k is a constant Qa ` a

f x F g x = xlim f x F xlim g x =AFB 3 A xlim Qa Qa Qa B ` a

` aC

` a

` a

4 A xlim f x A g x = xlim f x A xlim g x = AAB Qa Qa Qa B ` a

` aC

` a

` a

` a

` a B C ff xf lim ff xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f A f f f f f ` a = xQ a ` a = if B ≠ 0 5 A xlim Qa g x lim g x B xQ a w w w w w w w w w w w w w w w w w w w w w `w a

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w `w a

w w w w w w w w nw

n n q 6 A xlim f x =q lim f x = p A , Qa xQ a

w w w w w w w w nw

if p A is a real number

The uniqueness of a limit : If

lim f x = A

xQ a

then,

` a

lim f x = B

and

xQ a

` a

A = B

The Pinching Theorem :

If x is close to a but different from a, a function f always lies between two functions h and g. If, as x tends to a, both h and g tend to the same limit A, then f(x) also tends to A at x = a. Suppose, If then

h x ≤ f x ≤ g x ` a

lim h x = A xQ a ` a

` a

` a

and lim f x = A A xQ a ` a

for all x where 0 < |x @ a| < p and p>0, lim g x = A, xQ a ` a

LIMITS OF TRIGONOMETRIC FUNCTIONS The following trigonometric limits, which can be proven by the definition, is quite useful : lim sin x = sin c

lim cos x = cos c

xQ c

xQ c

sin xf f f f f f f f f f f f =1 lim f xQ 0 x

1f @ cos xf f f f f f f f f f f f f f f f f f f f f f f f f f f f lim f =0 xQ 0 x

sin 4x f f f f f f f f f f f f f f f f Example : Find lim f xQ0 x Solution : sin 4x f f f f f f f f f f f f f f f f f xQ 0 x f g sin 4x f f f f f f f f f f f f f f f f f = lim 4 xQ 0 4x lim

= 4 lim

4x Q 0

= 4 B1 =4

f

sin 4x f f f f f f f f f f f f f f f f f 4x

g

B

as x Q 0, we can write 4x Q 0

F

using lim

sin xf f f f f f f f f f f f f G =1 xQ 0 x

tan 3x f f f f f f f f f f f f f f f f f f f f f f f f f Example : Find lim f x Q 0 2x 2 + 5x

C

lim

tan 3x f f f f f f f f f f f f f f f f f f f f f f f f f f

xQ0

2x 2 + 5x

= lim

Ff G sin 3x f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f ` a

xQ0

= lim

x 2x + 5 A cos 3x

Ff G 3f sin 3x f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f ` a

xQ0

3 x 2x + 5 A cos 3x

F sin G 3x 1f 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Solution : = 3 lim f B B xQ0 3x 2x + 5 cos 3x

= 3 lim

F sin G 3x f f f f f f f f f f f f f f f f f

3x

xQ0

f gf g ` a 1f f f 1f f

=3 1 =

3f f f 5

5

1

B lim

Ff G 1f f f f f f f f f f f f f f f f f f f

xQ 0

2x + 5

B lim

Ff G 1f f f f f f f f f f f f f f f f f f

xQ 0

cos 3x

CONTINUITY While observing the graphs of different functions we note that some functions are discontinuous at some points. For example, the graph of x=

` a 1f πf f f f f f f f A The graph of f x = is discontinuous at x = 0. 2 x

f x = tan x is discontinuous at ` a

2 ` a xf @ 2x f f f f f f f f f f f f f f f f f f f f f , we see that the graph is discontinuous at x=2, but In the above graph of f x = f x @2

continuous at all other points.

Is there any other method, by which we know if the graph of a function is continuous or discontinuous at a point? The answer is, we can know it by using limits.

` a

Definition of Continuity : A function f x is continuous at a, ` a ` a f x = f a if and only if xlim Qa

We can also define continuity as : f is continuous at a if and only if, for each ε > 0 there exists δ > 0 such that ` a ` a if |x @ c| < δ then |f x @ f c | < ε A thus, f(x) will be continuous at a, 1. if f(a) is defined, ` a

2. if xlim f x exists and Qa lim f x = f a A

3. if

xQ a

` a

` a

A function f(x) is discontinuous at x = a, if any of the above conditions of continuity fails there. Geometrically it means that there is no gap, split or missing point in the graph of f(x) at a.

A function f is continuous in a closed interval [a, b] if it is continuous at every point in [a,b]. So, the function

f x = tan x is continuous in the interval ` a

πf f f f πf f f f as it is @ f , f 4 4

D

E

continuous at all the points in this interval, whereas it is not continuous in the interval F πf G f f f f 3π f f f f f f f f

4

,

πf f f f as it is discontinuous at x = f A 4 2

A function is said to be continuous if it is continuous at every point of its domain. So, f x = x 2 + 4 is a continuous function as it is continuous at every point in its domain (its ` a

domain is the set of real numbers R).

ONE-SIDED CONTINUITY: ` a

Definition :A function f x is continuous from left at a, if and only if

lim f x = f a x Q a@

` a

A function f x is continuous from right at a, if and only if

` a

` a

lim+ f x = f a

xQa

` a

` a

THEOREMS ON CONTINUITY The most important theorems of continuity are: If the functions

f and g are continuous at a, then

1 A f + g is continuous at a, 2 A αf is continuous at a for each real α, 3 A f g is continuous at a, ` a ff f f f 4A f is continuous at a provided g a ≠ 0 A g ` a

5 A If g is continuous at a and f is continuous at g a , then the composition of functions f ο g is continuous at a A

Two more theorems on continuity are : The Intermediate @ ValueB Theorem : C ` a ` a 6 A If f is continuous on a,b and C is a number between f a and f b , then there is at least one number c between a and b for which f c = C A ` a

The Maximun @ Minimum Theorem : B C 7 A If f is continuous on a,b , then f takes on both a maximum value M and a minimum B

C

value N on a,b A

Example : Determine the continuity of f x = 3x @ 5 at x = -3 ` a

a)

2 ` a xf @ 9f f f f f f f f f f f f f f f f f at x = -3 f x = f x+3

b) Solution : a

a

lim + f x = ` a

xQ@3

lim @ f x = ` a

xQ@3

lim + 3x @ 5 = 3 @ 3 @ 5 = @ 14

xQ@3

`

a

a

`

a

lim @ 3x @ 5 = 3 @ 3 @ 5 = @ 14

xQ@3

`

a

f @ 3 = 3 @ 3 @ 5 = @ 14 Hence, f is continuous at x = @ 3 A `

`

a

`

a

2 ` a ` a @ 9f xf f f f f f f f f f f f f f f f f b For f x = f we find that f @ 3 is undefined and does not exist A x+3 Hence, f is discontinuous at x = @ 3 A

` a

Example : Discuss the continuity of X 2 ^ ^ f @ 9f f f f f f f f f f f f f f f f f ` a \ xf f x =^ x @ 3 , x ^ Z

≠ 3 at x = 3. x =3

6,

Solution :

lim+ f x = lim+ ` a

` a` a xf +f 3f xf @ 3f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

x @3 similarly lim@ f x = 6 xQ 3

xQ 3

xQ 3

# lim f x = 6 xQ 3

` a

= lim+ x + 3 = 3 + 3 = 6 xQ 3

` a

`

and f 3 = 6 ` a ` a As lim f x = f 3 ; f is continuous at x = 3 A ` a

xQ 3

Example : Discuss the continuity of f x = ` a

6 + 3x, x 2 @ 4,

V

x< @ 2 at x = -2. x ≥@2

a

Solution : f @2 = @2 @4 =4@4 =0

` a

`

1

` a

2

a

a2

`

lim @ f x = lim @ 6 + 3x = 6 + 3 @ 2 = 0

xQ@2

` a

xQ@2

`

b

a

c `

`

a

lim + f x = lim + x 2 @ 4 = @ 2 @ 4 = 0

xQ@2

` a

xQ@2

Hence lim f x = 0 xQ@2

` a

as

a2

` a

lim + f x

xQ@2

= lim @ f x xQ@2

As lim f x = f @ 2 ; f is continuous at x = 3 A xQ@2

` a

`

a

` a