Limits and Continuity
Limit
The Idea of a Limit
We start with a number c and a function f defined at all numbers x near
c but not necessarily at c itself. In any case, whether or not f is defined
at c and, if so, how is totally irrelevant.
Now let L be some real number. We say that the limit of f (x) as x tends
to c is L and write
lim f ( x) = L
x→a
if the values of f ( x) approach L as x approaches a.
y = f ( x)
L
a
provided that (roughly speaking) as x approaches c, f(x) approaches L
or
(somewhat
more
precisely)
provided
that
f (x) is close to L for all x ≠ c which are close to c.
Example 1
Set f(x) = 4x + 5 and take c = 2. As x approaches 2, 4x approaches 8
and 4x + 5 approaches 8 + 5 = 13. We conclude that
Limits, Graphs, and Calculators
 x −1 
1. a) Use table of values to guess the value of lim  2 
x →1 x − 1


x −1
b) Use your calculator to draw the graph f ( x) = 2
x −1
and confirm your guess in a)
Graph 1
2. Find the following limits
 sin x 
a) lim 
by considering the values

x →0
 x 
x =±1, ±0.5, ±0.1, ±0.05, ±0.001. Thus the limit is 1.
sin x
Confirm this by ploting the graph of f ( x) =
x
Graph 2
( x ) by considering the values
b) lim sin π
x →0
(i) x =
±1, ± 1 , ± 1
, ±1
10
100
1000
(ii) x =
±1, ± 2 , ± 2
, ±2
3
103
1003
This shows the limit does not exist.
( x)
Confrim this by ploting the graph of f ( x) = sin π
Graph 3
c) Find
−3 x if x ≠ −2
lim f ( x) where f ( x) = 
x →−2
1 if x = −2
lim f ( x) = lim − 3 x
x →−2
6
= −3 lim x
x →−2
=−3(−2) =6
Note: f (-2) = 1
is not involved
x →−2
-2
3)
Use your calculator to evaluate the limits
a.
 4( x 2 − 4) 
lim 

x→2
−
x
2


b.
1, if x < 0
lim g ( x), where g ( x) = 
x →0
−1, if x ≥ 0
c.
1
lim f ( x), where f ( x) = 2
x →0
x
d.
 1+ x −1 
lim 

x →0
x


The ε -δ Definition of Limit
We say lim f ( x) = L if and only if
x→a
given a positive number ε , there exists a positive δ such that
if 0 <| x − a |< δ , then | f ( x) − L |< ε .
L+ε
L −ε
y = f ( x)
L
a
a −δ
a +δ
This means that if we are given a
small interval ( L − ε , L + ε ) centered at L,
then we can find a (small) interval (a − δ , a + δ )
such that for all x ≠ a in (a − δ , a + δ ),
f ( x) is in ( L − ε , L + ε ).
Examples
4. Show that lim(3 x + 4) =
10.
x→2
Let ε > 0 be given. We need to find a δ > 0 such that
if | x - 2 |< δ , then | (3 x + 4) − 10 |< ε .
But | (3 x + 4) − 10 |= | 3 x − 6 |= 3 | x − 2 |< ε
ε
if | x − 2 |<
3
So we choose δ =
ε
3
.
1
5. Show that lim = 1.
x →1 x
Let ε > 0 be given. We need to find a δ > 0 such that
if | x − 1|< δ , then | 1 − 1|< ε .
x
x −1 1
1
But |
| x − 1| .
− 1|
= |
=|
x
x
x
What do we do with the
x?
1
If we decide | x − 1|< , then 1 < x < 3 .
2
2
2
1
And so <2.
x
1
1
Thus |
| x − 1|< 2 | x − 1| .
− 1|
=
x
x
ε 1 
Now we choose δ = min  ,  .
3 2
1/2 1 3/2
One-Sided Limit
One-Sided Limits
The right-hand limit of f (x), as x approaches a,
equals L
written: lim+ f ( x) = L
x →a
if we can make the value f (x) arbitrarily close
to L by taking x to be sufficiently close to the
right of a.
y = f ( x)
L
a
The left-hand limit of f (x), as x approaches a,
equals M
written:
lim− f ( x) = M
x →a
if we can make the value f (x) arbitrarily close
to L by taking x to be sufficiently close to the
left of a.
y = f ( x)
M
a
Examples
Examples of One-Sided Limit
6. Given
 x 2 if x ≤ 3
f ( x) = 
2x if x > 3
Find lim+ f ( x)
x →3
f ( x) lim
=
lim=
2x 6
+
+
x →3
x →3
Find lim− f ( x)
x →3
2
lim=
f
(
x
)
lim
=
x
9
−
−
x →3
x →3
More Examples
 x + 1, if x > 0
7. Let f ( x) = 
 x − 1, if x ≤ 0.
a)
b)
Find the limits:
lim+ =
f ( x) lim+ ( x + 1)
x →0
x →0
= 0 +1 = 1
lim− =
f ( x) lim− ( x − 1) =0 − 1 =−1
x →0
x →0
x + 1) = 1 + 1 = 2
c) lim− =
f ( x) lim(
−
x →1
x →1
d) lim+ f (=
x)
x →1
lim(
x + 1)
+
x →1
= 1+1 = 2
A Theorem
lim f ( x)
x→a
if lim+ f ( x) L=
and lim− f ( x) L.
L if and only
=
x→a
x→a
This theorem is used to show a limit does not
exist.
For the function
 x + 1, if x > 0
f ( x) = 
 x − 1, if x ≤ 0.
lim f ( x) does not exist because lim+ f ( x) = 1 and lim− f ( x) = −1.
x →0
But
lim f ( x)
x →1
x →0
x →0
2 because
=
lim+ f ( x) 2=
and lim− f ( x) 2.
x →1
x →1
Limit Theorems
If c is any number,
lim f ( x) L=
and lim g ( x) M , then
=
x→a
a)
c)
e)
g)
i)
x→a
L−M
b) lim ( f ( x) − g ( x) ) =
lim ( f ( x) + g ( x) ) =
L+M
x→a
x→a
lim ( f ( x) ⋅ g ( x) ) =
L⋅M
(
x→a
lim ( c ⋅ f ( x) ) =
c⋅L
x→a
lim c = c
x→a
lim x n = a n
x→a
)
( x)
L , ( M ≠ 0)
lim f =
g ( x)
M
x→a
d)
f)
h)
j)
lim ( f ( x) ) = Ln
n
x→a
lim x = a
x→a
lim =
f ( x)
x→a
L , ( L > 0)
Examples Using Limit Rule
+ 1) lim x + lim1
Ex. lim ( x =
2
2
x →3
x →3
=
x →3
( lim x ) + lim1
2
x →3
x →3
= 32 + 1 = 10
lim ( 2 x − 1)
2 lim x − lim1
2x −1
x
1
→
x →1
x →1
Ex. lim
=
=
x →1 3 x + 5
lim ( 3 x + 5 )
3lim x + lim 5
x →1
x →1
=
x →1
2 −1 1
=
3+5 8
More Examples
1. Suppose lim f ( x) = 4 and lim g ( x) = −2. Find
x →3
x →3
a) lim ( f ( x) +=
g ( x) )
x →3
x →3
x →3
= 4 + (−2) = 2
b) lim ( f ( x) =
− g ( x) )
x →3
lim f ( x) + lim g ( x)
lim f ( x) − lim g ( x)
x →3
= 4 − (−2) = 6
x →3
2 f ( x) − lim g ( x)
 2 f ( x) − g ( x)  lim
x →3
x →3
c) lim 
=
=

x →3
lim f ( x) ⋅ lim g ( x)
 f ( x) g ( x) 
x →3
x →3
2 ⋅ 4 − (−2) −5
=
4 ⋅ (−2)
4
More Examples
a)
 x −3
lim 
=


x →9
 x −9 
 ( x − 3)( x + 3) 
lim 

x →9
 ( x − 9)( x + 3) 


x −9
=
= lim 

x →9 ( x − 9)( x + 3)


b)
 4 − x2 
 (2 − x)(2 + x) 
lim  2
 = lim 
x →−2 2 x + x 3

 x →−2  x 2 (2 + x) 
 2− x
= lim 

2
x →−2
 x 
2 − (−2) 4
=
= = 1
2
(−2)
4
 1  1
=
lim 
 6
x →9
+
x
3


L’ Hopital Rule (Limits)
Indeterminate Forms
Indeterminate forms occur when substitution in the limit
results in 0/0. In such cases either factor or rationalize the
expressions.
Ex.
x+5
Notice
x →−5 x 2 − 25
x+5
= lim
x →−5 ( x − 5 )( x + 5 )
lim
1
1
= lim
=
x →−5 ( x − 5 )
−10
0
form
0
Factor and cancel
common factors
The Squeezing Theorem
If f ( x) ≤ g ( x) ≤ h( x) when x is near a, and if
g ( x) = L
=
=
f ( x) lim
h( x) L, then lim
lim
x→a
x→a
x→a
( )
Example: Show that lim x 2 sin π = 0.
x
x →0
Note that we cannot use product rule because lim sin π
x →0
( )
( )
But − 1 ≤ sin π
( x ) DNE!
≤ 1 and so − x 2 ≤ x 2 sin π ≤ x 2 .
x
x
Since lim x 2 = lim(− x 2 ) = 0, we use the Squeezing Theorem to conclude
x →0
x →0
( x ) = 0.
lim x 2 sin π
x →0
Limits at Infinity
Limits at Infinity
1
lim n
For all n > 0, =
x →∞ x
1
lim
=
0
n
x →−∞ x
1
provided that n is defined.
x
5 + 1
3
+
2
3x + 5 x + 1
x
x
= lim
Ex. xlim
2
2 −4
→∞
x →∞
2 − 4x
x2
2
=
( x ) + lim ( 1 x ) = 3 + 0 + 0 = − 3
lim 3 + lim 5
x →∞
x →∞
( x)
lim 2
x →∞
Divide
2
by x
2
x →∞
− lim 4
x →∞
2
0−4
4
More Examples
 2 x3 − 3x 2 + 2 
1. lim  3

2
x →∞
−
−
+
x
x
x
100
1


 2 x3 3x 2 2
− 3 + 3

3
x
= lim  3 x 2 x
x →∞ x
 − x − 100 x + 1
 3
3
3
3
x
x
x
x
3 2


 2 − x + x3

= lim 

x →∞
1
100
1
 1− − 2 + 3 
x x
x 

2
= = 2
1






 4 x − 5 x + 21 
lim  3

x →∞ 7 x + 5 x 2 − 10 x + 1


2
2.


4 x 2 5 x 21
− 3+ 3


3
x
x
x
= lim  3

2
x →∞ 7 x
5
10
1
x
x


 3 + 3 − 3 + 3
x
x
x 
 x
21
 4 5
−
+
 x x 2 x3
= lim 
x →∞
5 10 1
7+ − 2 + 3
x x
x

0
=
7
=0





3.
 x2 + 2x − 4 
lim 

x →∞
+
12
x
31


 x2 2 x 4 
− 
 +
x x
= lim  x
x →∞
 12 x + 31 
 x
x 

4

 x+2− x 
= lim 

x →∞
31
 12 +

x 

∞+2
=
12
= ∞
4.

= lim 
x →∞


lim
x →∞
(
(
x +1 − x
2
x2 + 1 − x
1
)
)

x +1 + x 

x2 + 1 + x 

2
 x2 + 1 − x2 
= lim 

2
x →∞
 x +1 + x 


1
= lim 

2
x →∞
 x +1 + x 
1
1
=
= = 0
∞+∞ ∞
Infinite Limits
20
For all n > 0,
15
10
lim+
x →a
1
( x − a)
n
5
= ∞
2
-2
-4
-6
-8
-5
-10
-15
-20
40
lim−
x→a
30
1
( x − a)
n
= ∞ if n is even
20
10
-2
2
4
6
-10
-20
lim−
x→a
1
( x − a)
n
20
= −∞ if n is odd
15
10
5
-8
-6
-4
-2
2
-5
More Graphs
-10
-15
Examples
Find the limits
1.
2.
 3x 2 + 2 x + 1 
lim+ 

2
x →0
2x


 3+ 2 + 1 2 
x
x 
= lim+ 

x →0 
2



 2x +1 
 2x +1 
lim + 
 = lim
x →−3  2 x + 6  x →−3+  2( x + 3) 


40
20
-8
-6
-4
-2
2
-20
3+ ∞ + ∞
=
= ∞
2
= −∞
Limit and Trig Functions
From the graph of trigs functions
f ( x) sin
=
x and g ( x) cos x
1
1
0.5
0.5
-10
5
-5
10
-10
-0.5
-5
5
10
-0.5
-1
-1
we conclude that they are continuous everywhere
lim sin x sin
=
c and lim cos x cos c
x →c
x →c
Tangent and Secant
Tangent and secant are continuous everywhere in their
domain, which is the set of all real numbers
x ≠ ±π
2
, ± 3π
2
, ± 5π
2
, ± 7π
2
y = sec x
,
15
30
y = tan x
10
20
5
10
-6
-6
-4
-2
2
4
6
-4
-2
2
-5
-10
-10
-20
-15
-30
4
6
a) lim + sec x =
( 2)
x→ π
c)
lim
(
x → −3π
2
)
Examples
−∞ b) x→lim
π
tan x = −∞
+
e) lim− cot x = −∞
x →π
( 2)
d)
sec x
−
lim
(
x → −3π
2
)
−
= ∞
tan x =
f) lim tan x = 1
x →π
4
cos x 0
= = 0
g) lim cot x = x →lim
−3π ) sin x
1
(
x →( −3π )
2
2
∞
Limit and Exponential Functions
y
a , a >1
x
10
10
=
y a , 0 < a <1
x
8
8
6
6
4
4
2
2
-6
-4
-2
2
4
-6
6
-4
-2
2
4
-2
-2
The above graph confirm that exponential
functions are continuous everywhere.
lim a = a
x
x →c
c
6
Some examples
Example 1
lim
x →∞
(
x +x−x
2
)
Example 2
 π + tan x 
lim sin 

x →0
 tan x − 2sec x 
Example 3
lim [ x]
x→2
Limits continue …..
e −1
x −a
na n −1
lim
1,=
lim
x →0
x →∞ x − a
x
x
n
n
Evaluate the limit
Example 4
cos(2 x) − e
lim
x →0
sin 4 x
Example 5
1/ x 2
 sin x 
lim 

x →0
 x 
−2 x 2
Example 6
For all x
Find
lim f ( x)
x→4
Asymptotes
A line that a curve approaches, as it heads towards infinity.
The line y = L is called a horizontal asymptote
of the curve y = f ( x) if eihter
=
lim f ( x) L=
or lim f ( x) L.
x →∞
x →−∞
The line x = c is called a vertical asymptote
of the curve y = f ( x) if eihter
lim− f ( x) = ±∞ or lim+ f ( x) = ±∞.
x →c
x →c
Examples
Find the asymptotes of the graphs of the functions
x +1
1. f ( x) = 2
x −1
(i) lim− f ( x) = −∞
2
x →1
Therefore the line x = 1
is a vertical asymptote.
(ii) lim− f ( x) = +∞.
x →−1
Therefore the line x = −1
is a vertical asymptote.
(iii) lim f ( x) = 1.
x →∞
Therefore the line y = 1
is a horizonatl asymptote.
10
7.5
5
2.5
-4
-2
2
-2.5
-5
-7.5
-10
4
x −1
2. f ( x) = 2
x −1
 x −1 
(i) lim f ( x) = lim  2 
x →1
x →1
 x −1 
(iii) lim f ( x) = 0.
x →∞
Therefore the line y = 0


x −1
 1  1
=
= lim  =
lim
. is a horizonatl asymptote.



x →1
 ( x − 1)( x + 1)  x →1  x + 1  2
Therefore the line x = 1
is NOT a vertical asymptote.
(ii) lim+ f ( x) = +∞.
x →−1
Therefore the line x = −1
is a vertical asymptote.
10
7.5
5
2.5
-4
2
-2
-2.5
-5
-7.5
-10
4
Continuity
Continuity
A function f is continuous at the point x = a if
the following are true:
i ) f (a ) is defined
ii ) lim f ( x) exists
x →a
iii ) lim f ( x) = f (a )
x →a
f(a)
a
Examples
At which value(s) of x is the given function
discontinuous?
2
x −9
1. f ( x)= x + 2
2.
g ( x) =
x+3
Continuous everywhere
Continuous everywhere
except at x = −3
lim( x + 2) =a + 2
x→a
and so lim f ( x) = f (a)
x→a
g (−3) is undefined
4
6
2
4
-6
-4
-2
2
-2
2
-4
-4
-2
2
-2
4
-6
-8
-10
4
−1, if x ≤ 0
4. F ( x) = 
1, if x > 0
 x + 2, if x > 1
3. h( x) = 
1, if x ≤ 1
lim− h( x) = 1
x →1
and lim+ h( x)
x →1
lim+ F ( x) = 1 and lim− F ( x) = −1
x →0
=3
Thus h is not cont. at x=1.
h is continuous everywhere else
x →0
Thus F is not cont. at x = 0.
F is continuous everywhere else
5
3
4
3
2
2
1
1
-10
-2
2
4
-5
5
-1
-1
-2
-3
-2
-3
10
Continuous Functions
If f and g are continuous at x = a, then
f ± g , fg , and f
g
( g (a) ≠ 0 ) are continuous
at x = a
A polynomial function y = P(x) is continuous at
every point x.
A rational function R( x) = p( x) q( x) is continuous
at every point x in its domain.
Classification of Discontinuities
Removable discontinuity
Example 5
For the function defined by setting
1 + x 2 , x < 1

=
f ( x) =
3,
x 1
 4 − 2 x, x > 1

The point x0 = 1 is a removable discontinuity.
The left- and right-hand limits at 1 are as follows:
lim g(x) = lim (1 + x2) = 2,
x→1– x→1–
lim g(x) = lim (4 – 2x) = 2.
x → 1 + x → 1+
Thus, lim g(x) = 2. NOTE: It does not matter that g(1) ≠ 2.
x→1
 f ( x), x ≠ x0
g ( x) = 
 L, x = x0
Jump discontinuity
f ( x)
 x2 ,
x <1

0,
x 1
=

2
−
−
2
(
x
1)
,
x >1

In this case, a single limit does not
exist because the one-sided limits, L−
and L+, exist and are finite, but are
not equal:
L−
lim
=
f ( x) 1,=
L+ lim
=
f ( x) 2,
x →1−
x →1+
Essential discontinuity

 5 
x <1
sin

,

 x −1 

f ( x) =
0,
x 1
 1

,
x >1
 x − 1
For an essential discontinuity, only one of the two one-sided
limits needs not exist or be infinite.
Intermediate Value Theorem
If f is a continuous function on a closed interval [a, b]
and L is any number between f (a) and f (b), then there
is at least one number c in [a, b] such that f(c) = L.
y = f ( x)
f (b)
f (c) = L
f (a)
a c
b
Example
Given f ( x) = 3 x − 2 x − 5,
2
Show that f ( x) = 0 has a solution on [1, 2] .
f (1) =−4 < 0
f (2)= 3 > 0
f (x) is continuous (polynomial) and since f (1) < 0
and f (2) > 0, by the Intermediate Value Theorem
there exists a c on [1, 2] such that f (c) = 0.