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.)
limn-gon = circle
n
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!
Limit at Infinity (Cont.)
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.
Limit at Infinity (Cont.)
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
If c is any constant, then
1
2 
3
lim c = c and lim c = c
x+
lim
x+
lim
x-
c
p
x
c
n
x- x
=0, p > 0
=0, n  N
Example - 1
Evaluate : lim
x 
3x3 - 4x2 + 6x - 1
2x3 + x2 - 5x + 7
Solution :
We have lim
x 
3x3 - 4x2 + 6x - 1
2x3 + x2 - 5x + 7
4
6
1
+
x x2 x3
= lim
1 5
7
x 
2+ +
x x2
x3
3-
=
3-0+0-0
3
=
2+0-0+0
2
Dividing numerator and denominator by x3 


Example – 2
5x - 6
Evaluate : lim
4x2 + 9
x
Solution :
We have lim
x 
5x - 6
4x2 + 9
6

x 5 - 
x

= lim
x 
9
x 4 +

x2

=
5-0
4+0
=
5
2
5


= lim
x
4+
6
x
9
x2
Example - 3
Evaluate : lim
x
 x2 + x + 1 - x2 + 1 




Solution :
 x2 + x +1 - x2 +1 



x 
We have lim
= lim
x
=lim
x
 x2 + x + 1 - x2 + 1 



  2
  x + x + 1 + x2 + 1 

 x2 + x + 1 + x2 + 1  




x2 + x + 1 - x2 - 1
x2 + x + 1 + x2 + 1
Solution Cont.
=lim
x
x
x2 + x + 1 + x2 + 1
1
= lim
x 
=
1+
1
1
1
+
+ 1+
x x2
x2
1
1+ 0 + 0 + 1+ 0
1
1
=
=
1+1 2
Dividing numerator and denominator by x 
Example – 4
13 + 23 + ... + n3
Evaluate : lim
n4
n 
Solution :
We have lim
13 + 23 +...+ n3
n4
n




2
= lim
n2 n + 1
n 
= lim
n 
4n4

n2 n2 + 1 + 2n
4n4


2
n n + 1 
n3 = 

2





Solution (Cont.)

1
2
n 1 +
+ 
2
n

n
4
= lim
n 
4n4

1
1
2
= × lim 1 +
+ 
2
4 n  
n
n
1
1
= × 1 + 0 + 0  =
4
4
Continuity at a Point
Let f(x) be a real function and let x = a be any point in
its domain. Then f(x) is said to be continuous at x = a, if
lim f  x  exists and lim f  x  = f  a
x a
x a
 lim f  x  = lim f  x  = f  a
x  a-
x  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
lim f  x  exists and lim f  x  = f  a
x  a-
x  a-
f(x) is said to be right continuous at x = a if
lim f  x  exists and lim f  x  = f a
x  a+
x  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
2 
 3
f(x) is continuous on (a, b).
lim f  x  = f  a
x  a+
lim f  x  = f b 
x 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
f is continuous at x = a
 4
fg is continuous at x = a
5 
1
is continuous at x = a, f  a  0
f
6 
f
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.
3 The exponential function ax , a > 0 is
continuous everywhere.
(4) The logarithmic function is continuous in its domain.
(5) Inverse trigonometric functions are continuous in their domains.
(6) The composition of two continuous functions is a continuous function.
Example – 5
Determine the continuity of the function
 x cos x, x  0
f x = 
x=0
 0,
at x = 0.
Solution :
LHL at x = 0 =
lim f  x  = lim f 0 - h = lim | 0 - h | cos 0 - h 
x  0-
h0
h0
= lim h cosh = 0 ×1 = 0
h0
RHL at x = 0 =
lim f  x  = lim f 0 + h = lim | 0 + h | cos 0 + h
x  0+
h0
= lim h cosh = 0 ×1 = 0
h0
h0
Solution (Cont.)
and f 0 = 0
 lim f  x  = lim f  x   f 0 
x  0-
x  0+
So, f(x) is continuous at x = 0.
Example –6
Determine the continuity of the function
1
 2
x
sin
, x0

f x = 
x
 0,
x=0
at x = 0.
Solution :
LHL at x = 0 =
lim f  x  = lim x2sin
x  0-
x 0
1
x
= 0 × a finite oscillating number between - 1 and 1  = 0
RHL at x = 0 = lim f  x  = lim x2sin
x  0+
x 0
1
x
= 0 × a finite oscillating number between - 1 and 1  = 0
Solution (Cont.)
and f 0 = 0
 lim f  x  = lim f  x   f 0 
x  0-
x  0+
So, f(x) is continuous at x = 0.
Example – 7
Determine the continuity of the function

ex - 1
, if x  0

f  x  =  log 1 + 2x 

if x = 0
7,
at x = 0.
Solution :
ex - 1
ex - 1
x
= lim
×
LHL at x = 0  = lim- f  x  = lim
x
x  0 log 1 + 2x 
x 0
log 1 + 2x 
x 0
ex - 1
= lim
×
x
x 0
lim
x 0
1
log 1 + 2x 
2x
= 1×
×2
1
1
=
1× 2 2
Solution (Cont.)
ex - 1
ex - 1
x
= lim
×
RHL at x = 0  = lim+ f  x  = lim
x
x  0 log 1 + 2x 
x 0
log 1 + 2x 
x 0
ex - 1
= lim
×
x
x 0
lim
x 0
1
log 1 + 2x 
2x
= 1×
×2
and f 0 = 7
 lim f  x  = lim f  x   f 0 
x  0-
x  0+
So, f(x) is discontinuous at x = 0.
1
1
=
1× 2 2
Example – 8
Determine the value of the constant k so that the function
 sin5x
, if x  0

f  x  =  3x
k,
if x = 0
is continuous at x = 0.
Solution :
sin5x
sin5x 5
5 5
= lim
× = 1× =
3
3 3
x  0 3x
x  0 5x
LHL at x = 0 = lim f  x  = lim
x  0-
RHL at x = 0 =
sin5x
sin5x 5
5 5
= lim
× = 1× =
3
3 3
x  0 3x
x  0 5x
lim f  x  = lim
x  0+
Solution (Cont.)
and f 0 = k
The function f  x is continuous at x = 0.
 lim f  x  = lim f  x  = f 0 
x  0-

x  0+
5
5
= =k
3
3
k=
5
3
Example –9
Find the value of k if f  x  is continuous at x = 2, where
kx2 , if x  2
f x = 
if x > 2
3,
Solution :
LHL at x = 2 =
2
lim f  x  = lim kx2 = k 2  = 4k
x  2-
RHL at x = 2 =
2
x 2
lim f  x  = lim 3 = 3
x  2+
and f 2 = k 2  = 4k
x 2
Solution (Cont.)
The function f  x is continuous at x = 2.
 lim f  x  = lim f  x  = f 2 
x  2-
x  2+
 4k = 3 = 4k
k=
3
4
Example –10
Find the value of k if f  x  is continuous at x = 2, where
kx + 5,
f x = 
x - 1,
if x  2
if x > 2
Solution :
LHL at x = 2 =
RHL at x = 2 =
lim f  x  = lim kx + 5 = 2k + 5
x  2-
lim f  x  = lim  x - 1 = 2 - 1 = 1
x  2+
and f 2 = 2k + 5
x 2
x 2
Solution (Cont.)
The function f  x is continuous at x = 2.
 lim f  x  = lim f  x  = f 2 
x  2-
x  2+
 2k + 5 = 1 = 2k + 5
 2k + 5 = 1  2k = -4  k = -2
Thank you