3 Limits PDF
3 Limits PDF
3 Limits PDF
Objectives
- Know what left limits, right limits, and limits are.
- Know how to compute simple limits.
- Know what it means for a function to be continuous.
- Know what is the L Hopital′s rule.
1
What is a limit?
A limit is what happens when you get
closer and closer to a point without
actually reaching it.
Example: If 𝑓(𝑥) = 2𝑥 then as 𝑥 → 1,
𝑓 𝑥 → 2.
We write this as lim 𝑓 𝑥 = 2.
𝑥→1
2
Why are limits useful?
Many functions are not defined at a poin
but are well-behaved nearby.
𝑥 2 −1
Example: If 𝑓(𝑥) = then 𝑓 1 is
𝑥−1
undefined. However, as 𝑥 → 1, 𝑓 𝑥 → 2,
4
so lim 𝑓 𝑥 = 2 . 3
𝑥→1
2
1
f(x)0
-1
x 0 .9 .99 .999 .9999
-2
f(x) 0 1.9 1.99 1.999 1.999 -3
-4
-4 -3 -2 -1 0 1 2 3 4
x 3
Left Limits and Right Limits
𝑥
Consider 𝑓(𝑥) = . 𝑓 0 is undefined. As
𝑥
𝑥 → 0− , 𝑓(𝑥) = −1 4
x -1 -.1 -.01 -.001 -.0001 3
f(x) -1 -1 -1 -1 -1 2
1
As 𝑥 → 0+ , 𝑓(𝑥) = 1 f(x) 0
-1
x 1 .1 .01 .001 .0001
f(x) 1 1 1 1 1 -2
-3
-4
-4 -3 -2 -1 0 1 2 3 4
x
We write this as lim− 𝑓 𝑥 = −1 , lim+ 𝑓(𝑥) = 1
𝑥→0 𝑥→0
4
Limit Definition Summary
We say that lim− 𝑓 𝑥 = 𝐿 if 𝑓 𝑥 → 𝐿 as
𝑥→𝑎
𝑥 → 𝑎−
We say that lim+ 𝑓 𝑥 = 𝐿 if 𝑓 𝑥 → 𝐿 as
𝑥→𝑎
𝑥 → 𝑎+
If lim− 𝑓 𝑥 = lim+ 𝑓 𝑥 = 𝐿 (i.e. it
𝑥→𝑎 𝑥→𝑎
doesn’t matter which side x approaches
a from then we say that lim 𝑓 𝑥 = 𝐿
𝑥→𝑎
5
Absence of Limits
Limits can fail to exist in several ways
1. lim− 𝑓 𝑥 or lim+ 𝑓 𝑥 may not exist.
𝑥→𝑎 𝑥→𝑎
1
- Example: sin oscillates rapidly between
𝑥
+ − 1
0 and 1 as 𝑥 → 0 (or 0 ). Thus, lim+ sin
𝑥→0 𝑥
DNE (does not exist)
1
- Example: gets larger and larger as 𝑥 → 0+ .
𝑥
1
We write this as lim+ =∞
𝑥→0 𝑥
2. lim− 𝑓 𝑥 and lim+ 𝑓 𝑥 may both exist but
𝑥→𝑎 𝑥→𝑎
𝑥
have different values. Ex: 𝑓 𝑥 = near
|𝑥|
𝑥=0 6
Computing Limits
To compute lim 𝑓 𝑥 :
𝑥→𝑎
If nothing special happens at 𝑥 = 𝑎, just
compute 𝑓 𝑎 . Example: lim (3𝑥 − 1) = 5
𝑥→2
0
If plugging in 𝑥 = 𝑎 gives , factors can
0
often be cancelled when 𝑥 ≠ 𝑎.
Example:
𝑥 2 −4 (𝑥−2)(𝑥+2)
lim ( ) = lim ( ) = lim (𝑥 + 2) = 4
𝑥→2 𝑥 −2 𝑥→2 𝑥 −2 𝑥→2
7
Computing Limits
Continued
𝑎+𝑏 𝑎2 −𝑏 2
Useful trick: 𝑎 − 𝑏 = 𝑎 − 𝑏 ∙ =
𝑎+𝑏 𝑎+𝑏
𝑥+1−1
Example: What is lim ?
𝑥→0 𝑥
𝑥+1−1 𝑥+1−1 𝑥+1+1
lim = lim ∙
𝑥→0 𝑥 𝑥→0 𝑥 𝑥+1+1
𝑥 1 1
= lim = lim =
𝑥→0 𝑥( 𝑥 + 1 + 1) 𝑥→0 ( 𝑥 + 1 + 1) 2
8
Limits at Infinity
9
Computing Limits at ±∞
General strategy : figure out the
largest terms and ignore
everything else
3𝑥 2 −𝑥
Example: If 𝑓 𝑥 = , as
4𝑥 2 +2𝑥 −5
𝑥 → ∞ only the 3𝑥 2 in the
numerator and the 4𝑥 2 will really
3
matter, so lim 𝑓 𝑥 =
𝑥→∞ 4
10
Limit Laws
If lim 𝑓 𝑥 = L and lim 𝑔 𝑥 = 𝑀
𝑥→𝑎 𝑥→𝑎
then:
lim (𝑓 𝑥 + 𝑔(𝑥)) = L + M
𝑥→𝑎
lim (𝑓 𝑥 − 𝑔(𝑥)) = L − M
𝑥→𝑎
lim (𝑓 𝑥 𝑔(𝑥)) = LM
𝑥→𝑎
𝑓(𝑥) 𝐿
lim ( ) = (if 𝑀 ≠ 0)
𝑥→𝑎 𝑔(𝑥) 𝑀
Etc.
11
Continuity
Definition: 𝑓 𝑥 is continuous at a if
both 𝑓 𝑎 and lim 𝑓 𝑥 exist and are
𝑥→𝑎
equal.
Note: Polynomials are always
continuous everywhere. Most
functions we will be working with are
continuous almost everywhere.
12
Discontinuous functions
𝑓 𝑥 may fail to be continuous at 𝑥 = 𝑎
because:
1. lim 𝑓 𝑥 or 𝑓 𝑎 does not exist.
𝑥→𝑎
Example: If 𝑓 𝑥 = 𝑥 then lim 𝑓 𝑥 does
𝑥→0
not exist.
𝑥 2 −1
Example: If 𝑓 𝑥 = then 𝑓 1 is
𝑥−1
undefined.
2. lim 𝑓 𝑥 or 𝑓 𝑎 both exist but have
𝑥→𝑎
different values.
Example: If 𝑓 𝑥 = 𝑥 − 𝑥 then lim 𝑓 𝑥
𝑥→1
= 1 but 𝑓 1 = 0 13
L Hopital′s rule
Johann Bernoulli
1667 - 1748
Consider:
x2 4
lim
x 2 x 2
0
If we try to evaluate this by direct substitution, we get:
0
lim
x2 4
lim
x 2 x 2 lim x 2
4
x 2 x 2 x 2 x2 x 2
f x x2 4
lim lim
x a g x x 2 x 2
x 4
2 0.05
4
3
2
1
x
-3 -2 -1 1 2 3 0 1.95 2 2.05
x
0
-1
-2
-3
-0.05
-4
x2 -5
If we zoom in far enough,
the curves will appear as
straight lines.
f x x2 4
lim lim
x a g x x 2 x 2
0.05
df
As x2
f x becomes:
dg
g x
0 1.95 2 dx 2.05
x
df
df
dx
dg dg
-0.05 dx
f x x2 4
d 2
x 4
2x
lim lim lim dx lim 4
x a g x x 2 x 2 x 2 d
x 2
x 2 1
dx
L’Hôpital’s Rule:
f x
If lim is indeterminate, then:
x a g x
f x f x
lim lim
x a g x x a g x
Example:
0
1 0 0
0
0
The first one, , can be evaluated just like .
0
The others must be changed to fractions first.
1
lim x sin
x
x
This approaches 0
1
sin
lim x
This approaches
0
x 1
x 0
sin x
We already know that lim 1
x 0
x
but if we want to use L’Hôpital’s rule:
1 1 1
sin cos 2
x x x lim cos 1
lim lim
x 1 x 1 x
x
2
x x cos 0 1
1 1
lim
x 1
This is indeterminate form
x 1 ln x
x 1 ln x 0
lim
x 1 x 1 ln x
Now it is in the form
0
1
1 x
lim L’Hôpital’s rule applied once.
x 1 x 1
ln x
x
x 1 0
lim Fractions cleared. Still
x 1 x 1 x ln x
0
1 1
lim
x 1 ln x x 1 1
lim
x 1 1 1 ln x
x 1 ln x
lim
x 1 x 1 ln x
1
1 2
1 x
lim
x 1 x 1
ln x
x
x 1
lim
x 1 x 1 x ln x