Limit and continuity

Chapter 2Chapter 2
Limit and continuityLimit and continuity
 Tangent lines and length of the curveTangent lines and length of the curve
 The concept of limitThe concept of limit
 Computation of limitComputation of limit
 ContinuityContinuity
 Limit involving infinity (asymptotesLimit involving infinity (asymptotes

1)Tangent lines and the length of1)Tangent lines and the length of
the curvethe curve
 Tangent lineTangent line

 Lets zoom the graphLets zoom the graph

•Now we can see that the slope get closerNow we can see that the slope get closer
To get the correct tangent line of the graph at any point, we need toTo get the correct tangent line of the graph at any point, we need to
zoom the graph as can as possiblezoom the graph as can as possible

The length of curveThe length of curve
2 2
2 2
(4 1) (3 0)
+ (16 1) (6 3)
18 234 19.53
d = − + −
− + −
= + ≈
2 2
2
2 2
2 2
(0 4) (2 0)
(4 0) (4 2)
(16 4) (6 4)
21,1
d = − + −
+ − + −
+ − + −
≈
2 2
3
2 2
2 2
2 2
2 2
(1 4) (1 0)
+ (0 1) (2 1)
+ (2 0) (3.4 2)
+ (6 2) (4.4 3.4)
+ (16 6) (6 4.4)
21.25
d = − + −
− + −
− + −
− + −
− + −
≈

2)The concept of limit2)The concept of limit
 Example 1Example 1
 f is not defined at x=2f is not defined at x=2
 Example 2Example 2
 g is not defined at x=2g is not defined at x=2
2
( 4)
( )
( 2)
x
f x
x
−
=
−
2
( 5)
( )
( 2)
x
g x
x
−
=
−

Thus, we can conclude that the limit of f(x) when x approaches 2 is 4 and we writeThus, we can conclude that the limit of f(x) when x approaches 2 is 4 and we write
 We can see that when x get closer to 2We can see that when x get closer to 2
from left side, f(x) get closer to 4from left side, f(x) get closer to 4
1.91.9 3.93.9
1.991.99 3.993.99
1.9991.999 3.9993.999
1.99991.9999 3.99993.9999
2
( 4)
( )
( 2)
x
f x
x
−
=
−
x
2
( 4)
( )
( 2)
x
f x
x
−
=
−
2.12.1 4.14.1
2.012.01 4.014.01
2.0012.001 4.0014.001
2.00012.0001 4.00014.0001
x
2
( 4)
( )
( 2)
x
f x
x
−
=
−
We can see also that when x get closer to 2We can see also that when x get closer to 2
from right side, f(x) get closer to 4from right side, f(x) get closer to 4
2
lim ( ) 4
x
f x
→
=

Since the limits from the two sides are different, we conclude that the limit of g(x) when xSince the limits from the two sides are different, we conclude that the limit of g(x) when x
approaches 2 doesn’t existapproaches 2 doesn’t exist
 We can see that when x get closer to 2We can see that when x get closer to 2
from left side, g(x) increase very fastfrom left side, g(x) increase very fast
towardtoward
1.91.9 13.913.9
1.991.99 103.99103.99
1.9991.999 1003.9991003.999
1.99991.9999 10003.999910003.9999
2
( 5)
( )
( 2)
x
g x
x
−
=
−
x
2
( 5)
( )
( 2)
x
g x
x
−
=
−
2.12.1 -5.9-5.9
2.012.01 -95.99-95.99
2.0012.001 -995.999-995.999
2.00012.0001 -9995.9999-9995.9999
x
2
( 5)
( )
( 2)
x
g x
x
−
=
−
We can see also that when x get closer to 2We can see also that when x get closer to 2
from right side,g(x) decrease very fast towardfrom right side,g(x) decrease very fast toward
2
lim ( ) does not exist
x
g x
→
+∞
−∞

 Def:Def:
 A limit exist if and only if the two one-sided limit exist and are equalA limit exist if and only if the two one-sided limit exist and are equal
 for some L, if and only iffor some L, if and only if
 Exercise:Exercise:
 Find out why existFind out why exist
while does not existwhile does not exist
 Example:Example:
 Evaluate the below limit is it existEvaluate the below limit is it exist
lim ( )
x a
f x L
→
= lim ( ) lim ( )
x a x a
f x f x L+ −
→ →
= =
22
2
lim
4x
x
x→−
+
−
22
2
lim
4x
x
x→
+
−
0
| |
lim
x
x
x→
0 0
0 0
0 0 0
| |
lim lim 1
| |
lim lim 1
| | | | | |
lim lim lim does not exist
x x
x x
x x x
x x
x x
x x
x x
x x x
x x x
+ +
− −
+ −
→ →
→ →
→ → →
= =
−
= = −
≠ ⇒

3)Computation of the limit3)Computation of the limit
 Theorem1:Theorem1:
 ExampleExample evaluate the following limitsevaluate the following limits
suppose that lim ( ) and lim ( ) both exist and
let c be any constant, the following then apply
1) lim( . ( )) . lim ( )
2) lim( ( ) ( )) lim ( ) lim ( )
3) lim( ( ). ( )) lim (
x a x a
x a x a
x a x a x a
x a x a
f x f x
c f x c f x
f x g x f x g x
f x g x f x
→ →
→ →
→ → →
→ →
=
± = ±
= ). lim ( ),
lim ( )
( )
4) lim( ) /( lim ( ) 0)
( ) lim ( )
x a
x a
x a x a
x a
g x and
f x
f x
g x
g x g x
→
→
→ →
→
= ≠
2
20
2
2
(2 )( 1)
1) lim
2 3
4
2) lim
2
x
x
x x
x x
x
x
→
→
− −
+ +
−
−
 Theorem2:Theorem2:
For any Polynomial p(x) and any realFor any Polynomial p(x) and any real
number a,number a,
 Theorem 3:Theorem 3:
suppose that , thensuppose that , then
 Example:Example: evaluateevaluate
lim ( ) ( )
x a
p x p a
→
=
lim ( )
x a
f x L
→
=
lim ( ) lim ( ) nn n
x a x a
f x f x L
→ →
= =
5
2
2
7
2
1) lim 2 2
2
2) lim
2
x
x
x
x
x
→
→
−
−
−

 Theorem 4:Theorem 4:
 Example:Example: evaluate the following limitevaluate the following limit
 Theorem 5Theorem 5 (squeeze theorem)(squeeze theorem)
 Suppose thatSuppose that
for all x in some interval andfor all x in some interval and
for some number L, then alsofor some number L, then also
 Example:Example: evaluate the limitevaluate the limit
 ExampleExample: ( a limit of piecewise-: ( a limit of piecewise-
defined function)defined function)

1 1
1 1
1 1
For any real number , we have
1) lim sin sin
2) lim cos cos
3) lim
4) lim ln ln , for 0
5) lim sin sin for 1 1
6) lim cos cos for 1 1
7) lim tan tan for
8
x a
x a
x a
x a
x a
x a
x a
x a
a
x a
x a
e e
x a a
x a a
x a a
x a a
→
→
→
→
− −
→
− −
→
− −
→
=
=
=
= >
= − < <
= − < <
= − ∞ < < ∞
( )
)if p(x) is Ploy and lim ( )
lim ( ( ))
x p a
x a
f x L
f p x L
→
→
= ⇔
=
1
1/2
0
1) lim sin
2) lim .cot
x
x
x
x x
−
→
→
( ) ( ) ( )h x f x g x≤ ≤
( , )c b
lim ( ) lim ( ) / ( , )
x a x a
h x g x L a c d
→ →
= = ∈
lim ( )
x a
f x L
→
=
2
0
lim cos (1/ )
x
x x+
→
2
0
2cos 1 for 0
lim ( ), where ( )
4 for 0xx
x x x
f x f x
e x→
 + + <

− ≥

4)Continuity of Functions4)Continuity of Functions
DefinitionDefinition
A function f is continuous at a pointA function f is continuous at a point xx == aa ifif
is definedis defined1) ( )f a
2) lim ( ) exist and
3) lim ( ) ( )
x a
x a
f x
f x f a
→
→
=

So the function sin(x)
is continuous at x=0

So the function f(x) is
not continuous at
x=3

So the function f(x)
is not continuous at
x=0

Example ( )Where the function tan is continuous?y x=
Solution
( )
( )
( )
( )
( )
sin
The function tan is continuous whenever cos 0.
cos
Hence tan is continuous at , .
2
x
y x x
x
y x x n n
π
π
= = ≠
= ≠ + ∈¢

example ( ) 2
1
Where the function f sin is continuous?
1
φ
φ
 
=  ÷
− 
Solution ( ) 2
1
The function f sin is continuous at all points
1
where it takes finite values.
φ
φ
 
=  ÷
− 
2 2
1 1
If 1, is not finite, and sin is undefined.
1 1
φ
φ φ
 
= ±  ÷
− − 
2 2
1 1
If 1, is finite, and sin is defined and also finite.
1 1
φ
φ φ
 
≠ ±  ÷
− − 
2
1
Hence sin is continuous for 1.
1
φ
φ
 
≠ ± ÷
− 

( )
( )
0
0
A number for which an expression f either is undefined or
infinite is called a of the function f . The singularity is
said to be , if f can be defi
singularity
removab ned in such a wayle that
x x
x
0the function f becomes continous at .x x=
Problem 14
( )
( )
( )
2
0
0
0
Which of the following functions have removable
singularities at the indicated points?
2 8
a) f , 2
2
1
b) g , 1
1
1
c) h sin , 0
x x
x x
x
x
x x
x
t t t
t
− −
= = −
+
−
= =
−
 
= = ÷
 
Answer
Removable
Removable
Not removable

 Theorem 1Theorem 1
 All PolynomialAll Polynomial
are continuous everywhere, isare continuous everywhere, is
continuous for all x if n is odd, andcontinuous for all x if n is odd, and
continuous for all x 0 if n is even.continuous for all x 0 if n is even.
Also is continuous for allAlso is continuous for all
, and for, and for
 Suppose that and isSuppose that and is
continuous at , thencontinuous at , then
1
sin ,cos ,tan , x
x x x e−
n
x
≥
ln x
0x >
1 1
sin andcosx x− −
1 1x− < <
 Suppose that f(x) and g(x) areSuppose that f(x) and g(x) are
continuous at x=a, thencontinuous at x=a, then
 (f+g)(x) is continuous at x=a(f+g)(x) is continuous at x=a
 (f-g)(x) is continuous at x=a(f-g)(x) is continuous at x=a
 (f.g)(x) is continuous at x=a(f.g)(x) is continuous at x=a
 (f/g)(x) is continuous at x=a if(f/g)(x) is continuous at x=a if
g(a) 0g(a) 0≠
lim ( )
x a
g x L
¬
= f
L
lim ( ( )) (lim ( )) ( )
x a x a
f g x f g x f L
→ →
= =
 ExampleExample
 Determine whereDetermine where
is continuous?is continuous?
2
( ) cos( 3)f x x x= + −
 CorollaryCorollary
 Suppose that is continuous atSuppose that is continuous at
and is continuous atand is continuous at
, then, then
 is continuous atis continuous at
( )g x a
( )f x ( )g x
( )fog x a

Practice on continuityPractice on continuity
Exercise 11Exercise 11: explain why each function is: explain why each function is
discontinuous at the given point bydiscontinuous at the given point by
indicating which of the three condition inindicating which of the three condition in
definition are not metdefinition are not met
2
if 2
( ) 3 if 2
3 2 if 2
x x
f x x
x x
 <

= =
 − >

 Exercise 23,19Exercise 23,19:: find all discontinuity of f(x).find all discontinuity of f(x).
If the discontinuity is removable, introduceIf the discontinuity is removable, introduce
the new function that remove thethe new function that remove the
discontinuity:discontinuity:

2
3
2
3 1 if 1
1) ( ) 5 if 1 1
3 if 1
2) ( ) ln( )
x x
f x x x x
x x
f x x x
− < −

= + − < <

>
=
 Exercise 34Exercise 34: determine the value of a: determine the value of a
and b that make the given functionand b that make the given function
continuouscontinuous

1
2
1 if 0
( ) sin (x/2) if 0 2
if 2
x
ae x
f x x
x x b x
−
 + <

= ≤ ≤

− + >

 Theorem4Theorem4: (intermediate value theorem): (intermediate value theorem)
 Suppose that is continuous onSuppose that is continuous on
closed interval [a, b], and W is anyclosed interval [a, b], and W is any
number between f(a) and f(b). Then,number between f(a) and f(b). Then,
there is a numberthere is a number
 For whichFor which
( )f x
[ , ]c a b∈
( )f c W=
 corollary2corollary2: Suppose that f(x): Suppose that f(x)
is continuous on closed intervalis continuous on closed interval
(a, b), and f(a) and f(b) have(a, b), and f(a) and f(b) have
opposite signs (f(a).f(b)<0), soopposite signs (f(a).f(b)<0), so
there is at least one numberthere is at least one number
 for which f(c)=0for which f(c)=0
( , )c a b∈

Limit involving infinity, AsymptotesLimit involving infinity, Asymptotes
 Example 1:Example 1:
0
1
lim
x x→

Limit and continuity

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