LIMIT AND CONTINUITY
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LIMIT AND
CONTINUITY
3.1 Limit Function at One Point
An intuitive
Let
1
1)(
2
−
−=
x
xxf
This function is not defined at x = 1 ( dividing by 0).We still ask, what happening to f(x) as x approaches 1.
The following table shows some values of f(x) for x near 1
x
f(x)
0.9 0.99 0.999 1.11.011.0010.9999 1.00011
21.9 1.99 1.999 1.9999 2.0001 2.001 2.01 2.1
1
º2
x x
f(x)
f(x)
Geometrically
The result in the table and graphic suggest that the values of f(x) approaches 2 as x approaches 1.
We denote this by writing
21
1lim
2
1=
−
−
→ x
x
x
we read : the limit as x approaches 1 of is 2
1
12
−
−
x
x
Definition (intuitive) :
Means that when x is near but different from c, then f(x) is near L.
Lxfcx
=→
)(lim
853lim1
=+→
xx
Example 1:
a.
b. 2
)2)(12(lim
2
232lim
2
2
2 −
−+=
−
−−
→→ x
xx
x
xx
xx512lim
2=+=
→x
x
3
3
3
9lim
3
9lim
99 +
+
−
−=
−
−
→→ x
x
x
x
x
x
xx9
)3)(9(lim
9 −
+−=
→ x
xx
x
63lim9
=+=→
xx
c.
d. 0
lim sin(1/ )x
x→
If values of x approach 0 we get value of f(x):
x
)/1sin( x
/2 2/2 3/2 4/2 5/2 6/2 7/2 8/2
1 0 -1 0 1 0 -1 0
0
?
If x approaches 0, sin(1/x) does not approach to any number. Thus limit does not exist.
Right- and Left-Hand Limits (one-sided limits)
cx
If x approaches c from the left side (from number on which is smaller than c), we get the left-hand limit, denoted by
)(lim xfcx −→
)(lim xfcx +→
Right- and Left-Hand Limits (one-sided limits)
c xIf x approaches c from the right side (from number on which is greater than c), we get the right-hand limit, denoted by
Example 4
+
=
1,2
10,
0,
)(2
2
xx
xx
xx
xf
)(lim0
xfx→
)(lim1
xfx→
Example 5
Let
a. Evaluate
)(lim2
xfx→
d. Sketch graph of f(x)
c. Evaluate
b. Evaluate
2
0 0
0
0 0
lim ( ) lim 0lim ( ) 0
lim ( ) lim 0
x x
x
x x
f x xf x
f x x
− −
+ +
→ →
→
→ →
= =
== =
1 1
2 1 1 1
1 1
lim ( ) lim 1
lim ( ) lim ( ) lim ( )lim ( ) lim 2 3
x x
x x x
x x
f x x
f x does not exist because f x f xf x x
− −
− +
+ +
→ →
→ → →
→ →
= =
= + =
2
2 2lim ( ) lim 2 6x x
f x x→ →
= + =
c. The formula of f(x) does not change at x = 2, thus
b. Because the formula of f(x) change at x = 1, then we findthe one-sided limit at point x = 1.
Answer
a. Because the formula of f(x) change at x = 0, then we findthe one-sided limit at point x = 0.
d.
For x 02)( xxf =
Graph: parabola
For 0 < x < 1
f(x) = x
Graph: straight line
For x 1 22)( xxf +=
Graph: parabola
at x = 1 limit does not exist
22)( xxf +=
1
3
º
2. Find the value of c so that
−−
−−=
1;
1;3)(
2 xifcx
xifcxxf
has a limit at x = -1
Solution :
f(x) has a limit at x = -1 if the limit from the right = the limit from the left
)(lim1
xfx −−→
ccxx
+=−=−−→
33lim1
)(lim1
xfx +−→
ccxx
−=−=+−→
1lim 2
1
limit exist if 3 + c = 1 - c
C = -1
)(lim3
xfx −→
)(lim1
xfx −→
)(lim1
xfx→
A. For the function f graphed below
Find: f(-3)
f(-1)
f(1)
1.
2.
3.
4.
5.
6.
EXERCISES
B.
+−
+=
1,2
1,1)(
2
2
xxx
xxxf
1 1lim ( ) lim ( )x x
f x and f x− +→ →
1lim ( )x
f x→
xxxg 32)( −−=
x
g x
→ −2
lim ( )
x
g x
→+
2
lim ( )
xg x
→2lim ( )
2
2)(
−
−=
x
xxf
x
f x
→ −2
lim ( )
x
f x
→ +2
lim ( )x
f x→2
lim ( )
1. Let
a. Evaluate
b. Does exist? if it limit exist evaluate
2. Let , evaluate
3. Let , evaluate
a. b. c.
a. b. c.
1lim ( )x
f x→
lim ( ) lim ( )andx a x a
f x L g x G→ →
= =
1. lim ( ) ( ) lim ( ) lim ( )
2. lim ( ) ( ) lim ( ) lim ( )
lim ( )( )
3. lim , 0( ) lim ( )
4. lim( ( )) (lim ( )) ,
5. lim ( ) lim ( )
x a x a x a
x a x a x a
x a
x ax a
n n
x a x a
n nx a x a
f x g x f x g x L G
f x g x f x g x LG
f xf x L
if Gg x g x G
f x f x n positive in teger
f x f x
→ → →
→ → →
→
→→
→ →
→ →
= =
= =
= =
=
= 0n L provide L if n is even=
Theorem
Let
then
222 )1(1
1sin)1()1( −
−−−− x
xxx
)()()( xhxgxf
lim ( ) , lim ( )x c x c
f x L h x L→ →
= =
Lxgcx
=→
)(lim
2
1
1lim ( 1) sin
1xx
x→−
−
The Squeeze Theorem (Sandwich)
Let for all x in some open interval containing the point c and
then
Example 6: Evaluate
Because 1)1
1sin(1
−−
x
and
01
1sin)1(lim 2
1=
−−
→ xx
xthen
2 2
1 1lim ( 1) 0, lim( 1) 0x x
x x→ →− − = − =
0 4 0
0 0
0 2 0
sin 4 sin 4 sin 4.4 lim .4 4 lim
sin 4 44 4 4lim lim 2tan 2 tan 2 tan 2tan 2 2
.2 lim .2 2 lim2 2 2
x x
x x
x x
x x x
x x x xx x xx
x x x
→ →
→ →
→ →
= = = = =
x → 0 equivalent with 4x → 0
Example 7 :
Exercises
t
t
t sin1
coslim
2
0 +→
t
tt
t sec2
sincotlim
0
→
t
t
t 2
3tanlim
2
0→
Evaluate:
1.
2.
3.
4.x
x
x 2sin
tanlim
0→
The Precise Definition of a Limit
• In general
Infinite Limits and Limits at Infinity
a. Infinite Limits
lim ( ) 0 lim ( ) 0 ,x a x a
f x L g x then→ →
= =Let and =→ )(
)(lim
xg
xf
ax
( ) , 0 ( ) 0i if L g x from the upward+ →and
( ) , 0 ( ) 0andii if L g x from downward− →
( ) , 0 ( ) 0andiii if L g x from downward+ →
( ) , 0 ( ) 0andiv if L g x from upward− →
Remark : g(x) → 0 from upward mean g(x) approaches 0 from positive value of g(x).
g(x) → 0 from downward mean g(x) approaches 0 from negative value of g(x).
Example 8: Evaluate
1
1lim
2
1 −
+−→ x
x
x
a.x
x
x sinlim
+→b. c.
Solution
a. 021lim 2
1=+
−→x
x
, g(x) = x - 1 approaches 0 from the downward
thus −=−
+−→ 1
1lim
2
1 x
x
x
b.2
1lim 1 2 0
xx
−→ −+ = , approaches 0 from the upward1)( 2 −= xxg
thus
2
21
1lim
1x
x
x−→ −
+= +
−
1
1lim
2
2
1 −
+−−→ x
x
x
c.
0lim =+→
xx
and
f(x)=sinx
x
If x approaches from the right then sin(x) approaches 0 from downward
−=+→ x
x
x sinlim
thus
Because
Limits at Infinity
Example 9 : Evaluate42
52lim
2
2
+
++
→ x
xx
x
Solution
)2(
)1(lim
2
2
42
522
x
xx
x x
x
+
++=
→42
52lim
2
2
+
++
→ x
xx
x
2
2
42
521
lim
x
xxx
+
++
=→
= 1/2
Example 10: Evaluate
42
52lim
2 +
+
−→ x
x
x
42
52lim
2 +
+
−→ x
x
x
Solution
)2(
)(lim
2
2
42
522
x
xx
x x
x
+
+=
−→ )2(
)(lim
2
2
4
52
x
xx
x +
+=
−→
= 0
Example 11 : Evaluate xxxx
+++−→
3lim 2
Solution:
If x → , It is form ( ) −
xxxx
+++−→
3lim 2
)3
3(3lim
2
22
xxx
xxxxxx
x −++
−+++++=
−→
xxx
xxx
x −++
−++=
−→ 3
3lim
2
22
xxx
x
x −++
+=
−→ 3
3lim
2
xx
x
xx
x
x −++
+=
−→ )1(
)1(lim
2
312
3
xx
x
xx
x
x −++−
+=
−→2
31
3
1
)1(lim
( )
( ) 22
3 3
3131
1 1 1 0 1lim lim
21 0 0 11 11 1
x x
x xx xx x
x
x→− →−
+ + += = = = −
− + + −− + + −− + + −
Exercise
.
Evaluatelim
x
x
x→ +
+
−3
3
3
limx x→ + −2
2
3
4
)1(lim xxx
−−→
limx
x
x→ +1 2
1
1lim
2
−
+
−→ x
x
x
limx
x x
x→
+
+
2
1
.
1.
2.
3.
4.
5.
6.
21
2 17. lim
2x
x
x x−→
−
+ −
20
tan8. lim
3x
x
x x→ −
2 2 59. lim
2 5x
x x
x→−
− +
+
2
2 310. lim
2x
x
x x→−
−
− −
Continuity
A Function f(x) is said continuous at x = c if
(i) f(c) is defined
(ii)
(iii) )()(lim cfxfcx
=→
If one or more of these 3 conditions fails, then f is discontinuous at c and cis a point of discontinuity of f
c
(i)
ºf(c) does not exist
f is not continuous at c
existxfcx
)(lim→
c
(ii)
1L
2L Left-hand limit ≠ right-hand limit ,Thus limit does not exist at x = c
f is not continuous at x = c
(iii)
c
●
º
f(c)f(c) is defined
)(lim xfcx→
L exist
f is not continuous at x = c
but lim ( ) (c)x c
f x f→
(iv)
c
f(c)
f(c) is defined
)(lim xfcx→
exist
)()(lim cfxfcx
=→
f(x) is continuous x=c
Removable discontinuity
For case (i) discontinuity can be removed by define f(c) = limit of function at c
Example
Determine whether f ,g ,and h are continuous at x = 2
2
4)(
2
−
−=
x
xxf
2 4, if 2
g( ) 2
3 , if 2
xx
x x
x
−
= − =
a. b.2
1, if 2h( )
1, if 2
x xx
x x
+ =
− c.
Solution:
a. f is not defined at x = 2 (0/0) f is not continuous at x = 2
b. g(2) = 3,
,42lim)2(
)2)(2(lim
2
4lim
22
2
2=+=
−
+−=
−
−
→→→x
x
xx
x
x
xxx
2lim g( ) (2).x
x g→
g is not continuous at x=2
c. 2h(2) 2 1 3= − =
2 2lim h( ) lim 1 3x x
x x− −→ →
= + =
2
2 2lim h( ) lim 1 3x x
x x+ +→ →
= − =2
lim h( ) 3x
x→
=
2lim h( ) (2)x
x h→
=
h is continuous at x = 2
Continuity From the Left and the Right
A function f(x) is called continuous from the left at x =a if
)()(lim afxfax
=−→
)()(lim afxfax
=+→
f(x) is continuous at x = a if f(x) is continuous from the left and right
Example : Find the value of a so that f(x) will be continuous at x = 2
−
+=
2;1
2;)(
2 xifax
xifaxxf
A function f(x) is called continuous from the right at x = a if
Solution :
f(x) is continuous at x = 2, if :
1. f is continuous from the left at x = 2
)2()(lim2
fxfx
=−→
aaxxfxx
+=+=−− →→
2lim)(lim22
2(2) (2) 1 4 1f a a= − = −
2 + a = 4a – 1-3a = -3
a = 1
2. f is continuous from the right at x = 2
)2()(lim2
fxfx
=+→
141lim)(lim 2
22−=−=
++ →→aaxxf
xx
trivial
2(2) (2) 1 4 1f a a= − = −
1. Let
−+
−−=
1,22
1,1)(
2
xx
xxxf
Determine whether f is continuous at x = -1
Problems
2. Find the value of a + 2b so that f(x) will be continuous everywhere
+
+
=
2,3
21,
1,1
)(
xx
xbax
xx
xf
3. Find the values of a and b so that f(x) will be continuous at x = 2
−
−
−+=
2,42
2,2
4)(
2
xx
xx
bxaxxf
Continuity on a Closed Interval
• A function f(x) is said to be continuous on closed interval [a,b] if
1. f(x) is continuous on ( a,b )
2. f(x) is continuous from the right at x = a
3. f(x) is continuous from the left at x = b
Theorem A
• Polynomials are continuous function
• A rational function is continuous everywhere except at the points where the denominator is zero
Theorem B
• Let , then
• f(x) is continuous everywhere if n is odd
• f(x) is continuous for positive number if n is even
Example: In which x s.t. is continuous ?
From theorem, f(x) is continuous for x – 4 > 0 or x > 4.
Thus f(x) is continuous at [4, )
n xxf =)(
4)( −= xxf
)4(04lim)(lim44
fxxfxx
==−=++ →→
f(x) is continuous from the right at x = 4
Example
• Show that there is a root of the equation 𝑥3 − 𝑥 −1 = 0 between 1 and 2
f xx x
x( ) =
+
+
2 3
3
f xx
x( ) =
−
−
2
3
4
8
f xx
x( )
| |=
−
−
2
2
94
1)(
2 −−
−=
x
xxf
24)( xxxf −=
A. Find the points of discontinuity, if any
B. Find the points where f(x) are continuous
Problems
1.
2.
3.
1.
2.
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