Mathematics

Mathematics

Session

Functions, Limits and Continuity -3

Session Objectives

Limit at Infinity

Continuity at a Point

Continuity Over an Open/Closed Interval

Sum, Product and Quotient of Continuous Functions

Continuity of Special Functions

Limit at Infinity

A GEOMETRIC EXAMPLE:Let's look at a polygon inscribed in a circle... If we increase the number of sides of the polygon, what can you say about the polygon with respect to the circle?

As the number of sides of the polygon increase, the polygon is getting closer and closer to becoming the circle! If we refer to the polygon as an n-gon, where n is the number of sides,Then we can write

Limit at Infinity (Cont.)

n

lim n- gon = circle

The n-gon never really gets to be the circle, but it will get very close! So close, in fact, that, for all practical purposes, it may as well be the circle. That's what limits are all about!


A GRAPHICAL EXAMPLE:

Now, let's look at the graph of f(x)=1/x and see what happens!

Let's look at the blue arrow first. As x gets really, really big, the graph gets closer and closer to the x-axis which has a height of 0. So, as x approaches infinity, f(x) is approaching 0. This is called a limit at infinity.


Now let's look at the green arrow... What is happening to the graph as x gets really, really small? Yes, the graph is again getting closer and closer to the x-axis (which is 0.) It's just coming in from below this time.

Some Results

x + x -

I f c is any constant, then

1 lim c=c and lim c=c

px +

c2 lim =0, p > 0

x

nx -

c3 lim =0, n N

x

Example - 1

3 2

3 2x

3x - 4x + 6x - 1Evaluate : lim

2x + x - 5x + 7

3 2

3 2x

3x - 4x +6x - 1We have lim

2x + x - 5x +7

2 3 3

x2 3

4 6 13 - + -

x x x= lim Dividing numerator and denominator by x1 5 7

2 + - +x x x

3 - 0 + 0 - 0 3= =

2 + 0 - 0 + 0 2

Solution :

Example – 2

2x

5x - 6Evaluate : lim

4x + 9

2x

5x - 6We have lim

4x +9

x x

22

6 6x 5 - 5 -x x= lim = lim99 4 +x 4 +xx

5 - 0 5= =

24 + 0

Solution :

Example - 3

2 2

xEvaluate : lim x + x +1 - x +1

2 2

xWe have lim x + x +1 - x +1

2 2

2 2

x 2 2

x + x +1 - x +1=lim x + x +1 + x +1

x + x +1 + x +1

2 2

2 2x

x + x +1 - x - 1=lim

x + x +1 + x +1

Solution :

Solution Cont.

2 2x

x=lim

x + x +1 + x +1

x

2 2

1=lim Dividing numerator and denominator by x

1 1 11+ + + 1+

x x x

1=

1+ 0 + 0 + 1+ 0

1 1= =

1+1 2

Example – 4

3 3 3

4n

1 + 2 +... + nEvaluate : lim

n

3 3 3

4n

1 +2 +...+nWe have lim

n

2223

4n

n n +1 n n +1=lim n =

24n

2 2

4n

n n +1 + 2n=lim

4n

Solution :

Solution (Cont.)

1 1= × 1 + 0 + 0 =

4 4

42

4n

1 2n 1 + +

nn=lim4n

2n

1 1 2= × lim 1 + +

4 nn

Continuity at a Point

Let f(x) be a real function and let x = a be any point inits domain. Then f(x) is said to be continuous at x = a, if

x a x alim f x exists and lim f x = f a

- +x a x a

lim f x = lim f x = f a

If f(x) is not continuous at x = a, then it is said to be discontinuous at x = a.

Left and Right Continuity

f(x) is said to be left continuous at x = a if

- -x a x a

lim f x exists and lim f x = f a

f(x) is said to be right continuous at x = a if

+ +x a x a

lim f x exists and lim f x = f a

Continuity Over an Open/Closed Interval

f(x) is said to be continuous on (a, b) if

f(x) is continuous at every point on (a, b).

f(x) is said to be continuous on [a, b] if

1 f(x) is continuous on (a, b).

+x a

2 lim f x = f a

-x b

3 lim f x = f b

Sum, Product and Quotient of Continuous Functions

Let f and g be continuous at x = a, and let be a real number, then

1 f + g is continuous at x = a

2 f - g is continuous at x = a

3 fi s continuous at x = a

4 fg is continuous at x = a

15 is continuous at x = a, f a 0

f

f6 is continuous at x = a, g a 0

g

Continuity of Special Functions

(1) A polynomial function is continuous everywhere.

(2) Trigonometric functions are continuous in their respective domains.

(5) Inverse trigonometric functions are continuous in their domains.

(4) The logarithmic function is continuous in its domain.

(6) The composition of two continuous functions is a continuous function.

x3 The exponential function a , a > 0 is continuous everywhere.

Example – 5

Determine the continuity of the function

x cos x, x 0f x = at x = 0.

0, x = 0

- h 0 h 0x 0

Solution :

LHL at x = 0 = lim f x = lim f 0 - h = lim | 0 - h | cos 0 - h

h 0= lim h cosh = 0 ×1 = 0

+ h 0 h 0x 0

RHL at x = 0 = lim f x = lim f 0 + h = lim | 0 + h | cos 0 + h

h 0= lim h cosh = 0 ×1 = 0

Solution (Cont.)

and f 0 = 0

- +x 0 x 0

lim f x = lim f x f 0

So, f(x) is continuous at x = 0.

Example –6

2


1x sin , x 0

f x = at x = 0.x0, x = 0

-

2

x 0x 0

Solution :

1LHL at x = 0 = lim f x = lim x sin

x

= 0 × a finite oscillating number between - 1 and 1 = 0

+

2

x 0x 0

1RHL at x = 0 = lim f x = lim x sin

x

= 0 × a finite oscillating number between - 1 and 1 = 0

Solution (Cont.)

and f 0 = 0

- +x 0 x 0


So, f(x) is continuous at x = 0.

Example – 7

x


e - 1, if x 0

f x = at x = 0.log 1 + 2x

7, if x = 0

-

x x

x 0 x 0x 0

Solution :

e - 1 e - 1 xLHL at x = 0 = lim f x = lim = lim ×

xlog 1 + 2x log 1 + 2x

x

x 0

x 0

e - 1 1 1 1= lim × = 1× =

x 1×2 2log 1 + 2xlim ×2

2x

Solution (Cont.)

+

x x

x 0 x 0x 0

e - 1 e - 1 xRHL at x = 0 = lim f x = lim = lim ×

xlog 1+ 2x log 1+ 2x

x

x 0

x 0

e - 1 1 1 1= lim × = 1× =

x 1×2 2log 1 + 2xlim ×2

2x

and f 0 = 7

- +x 0 x 0


So, f(x) is discontinuous at x = 0.

Example – 8

Determine the value of the constant k so that the function

sin5x, if x 0

f x = is continuous at x = 0.3xk, if x = 0

- x 0 x 0x 0

Solution :

sin5x sin5x 5 5 5LHL at x = 0 = lim f x = lim = lim × = 1× =

3x 5x 3 3 3

+ x 0 x 0x 0

sin5x sin5x 5 5 5RHL at x = 0 = lim f x = lim = lim × = 1× =

3x 5x 3 3 3

Solution (Cont.)

and f 0 = k

The function f x is continuous at x = 0.

- +x 0 x 0

lim f x = lim f x = f 0

5 5= = k

3 3

5k =

3

Example –9

2

Find the value of k iff x is continuous at x = 2, where

kx , if x 2f x =3, if x > 2

-

22

x 2x 2

Solution :

LHL at x = 2 = lim f x = lim kx = k 2 = 4k

+ x 2x 2

RHL at x = 2 = lim f x = lim 3 = 3

2and f 2 = k 2 = 4k

Solution (Cont.)


- +x 2 x 2


4k = 3 = 4k

3k =

4

Example –10

Find the value of k iff x is continuous at x = 2, where

kx + 5, if x 2f x =

x - 1, if x > 2

- x 2x 2

Solution :

LHL at x = 2 = lim f x = lim kx + 5 = 2k + 5

+ x 2x 2

RHL at x = 2 = lim f x = lim x - 1 = 2 - 1 = 1

and f 2 = 2k + 5

Solution (Cont.)


- +x 2 x 2


2k + 5 = 1 = 2k + 5

2k + 5 = 1 2k = -4 k = -2

Thank you

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