Limits at Infinity
and Infinite Limits
more examples of limits
Motivation:
handling infinite variable
and infinite function
2
Question. Can we describe
in mathematics:
(1) infinite value of variable?
(2) infinite value of function?
f(x)=1/x
1
O
2
Question. Can we describe
in mathematics:
(1) infinite value of variable?
f(x)=1/x
1
O
(2) infinite value of function?
Application: horizontal and vertical asymptotes
Limits at infinity
infinite value of variable
Definition. L = lim f (x)
x→∞
⇔ ∀ε > 0 ∃N > 0 / x > N ⇒ |f (x) − L| < ε
Definition. L = lim f (x)
x→∞
⇔ ∀ε > 0 ∃N > 0 / x > N ⇒ |f (x) − L| < ε
5x + 1
Example: lim
=5
x→∞ x − 2
⇔ ∀ε > 0 ∃N > 0 / x > N
5x + 1
⇒ − 5 < ε
x−2
(in particular, | 5x+1
x−2 − 5| =
11
|x−2|
< ε if N = 2 + 11
ε )
Properties: Let lim f (x) and lim g(x) exist,
x→∞
x→∞
lim [f (x) + g(x)] = lim f (x) + lim g(x)
x→∞
x→∞
x→∞
lim [f (x) − g(x)] = lim f (x) − lim g(x)
x→∞
x→∞
x→∞
lim [f (x) · g(x)] = [ lim f (x)] · [ lim g(x)]
x→∞
x→∞
x→∞
lim [f (x)/g(x)] = [ lim f (x)]/[ lim g(x)],
x→∞
x→∞
x→∞
lim g(x) 6= 0
x→∞
q
p
n
lim g(x) = n lim g(x), lim g(x) > 0, n even
x→∞
x→∞
x→∞
EXAMPLES
EXAMPLE 1. Evaluate limit
1
lim
x→∞ x
EXAMPLE 1. Evaluate limit
1
lim
x→∞ x
As variable x gets larger, 1/x gets smaller because
1 is being divided by a laaaaaaaarge number:
1
1
x = 1010,
= 10
x 10
EXAMPLE 1. Evaluate limit
1
lim
x→∞ x
As variable x gets larger, 1/x gets smaller because
1 is being divided by a laaaaaaaarge number:
1
1
x = 1010,
= 10
x 10
The limit is 0.
1
lim
= 0.
x→∞ x
EXAMPLE 2. Limits
1
lim n = 0
x→∞ x
1
lim √
=0
n
x→∞
x
EXAMPLE 2. Limits
1
lim n = 0
x→∞ x
1
lim √
=0
n
x→∞
x
By the same argument
1
lim
= 0,
x→∞ x − 100
1
=0
lim √
n
x→∞
x − 10000
EXAMPLE 3. Evaluate limit
1
lim √
x→∞ n x2 + x − C
EXAMPLE 3. Evaluate limit
1
lim √
x→∞ n x2 + x − C
Recall the graph of
2
y =x +x−C
Observation:
y= x 2+ x - C
x→∞⇒y→∞
Value of C does not matter.
The answer is 0.
O
-1/4-C
EXAMPLE 4. Evaluate limit (not evoke graphs)
1
lim 2
x→∞ x − x − 1
EXAMPLE 4. Evaluate limit (not evoke graphs)
1
lim 2
x→∞ x − x − 1
Attention: indeterminacy ∞ − ∞
1
= lim
1
1
x→∞
2
x 1− − 2
x x
1
= lim 2 lim
x→∞ x x→∞
1
=0·1=0
1
1
1− − 2
x x
EXAMPLE 5. Evaluate limit
5x3 − 2x2 − 1
lim
x→∞
x3 − x + 1
EXAMPLE 5. Evaluate limit
5x3 − 2x2 − 1
lim
x→∞
x3 − x + 1
Standard trick: divide by the highest degree of x
1
5x3 2x2
2
1
−
−
5
−
−
3
3
3
3
x
x
x
x
x
= lim
= lim
3
1
1
x→∞
x→∞
x
x
1
1− 2+ 3
−
+
x
x
x3 x3 x3
5−0−0
=
=5
1−0+0
EXAMPLE 6. Evaluate limit
lim p
x→∞
x−2
2x2 − x + 1
EXAMPLE 6. Evaluate limit
lim p
x→∞
x−2
2x2 − x + 1
Standard trick: divide by the highest degree of x
x 2
x
2
−
− 3
= lim √ x x
= lim r x x
x→∞
2x2 − x + 1 x→∞
2x2 x
1
√
−
+
x2
x2 x2
x2
1−0
1
=√
=√ .
2−0+0
2
Indeterminacy:
unsuitable breaking in limits
Attention: Indeterminacy
2
3
[x + 1]
∞
[x − 1]
lim 3
lim 2
x→∞ [x − 3]
x→1 [x − 1]
∞
0
0
Wrong breaking! Limit laws do not apply!
lim[1 − cos x][cot x] ( 0 · ∞ )
x→0
Substitution is undefined!
p
p
lim [ x2 + x] − [ x2 − x] ( ∞ − ∞ )
x→∞
Rewrite first!
∞
Indeterminacy :
∞
[x2 + 1]
lim 3
x→∞ [x − 3]
∞
Indeterminacy :
∞
[x2 + 1]
lim 3
x→∞ [x − 3]
division by highest exponent of x:
x2
1
+ 3 0+0
3
x
x
lim 3
=
=0
x→∞ x
1−0
3
− 3
3
x
x
0
Indeterminacy :
0
3
[x − 1]
lim 2
x→1 [x − 1]
0
Indeterminacy :
0
3
[x − 1]
lim 2
x→1 [x − 1]
factoring and re-grouping:
2
2
(x − 1)(x + x + 1)
(x − 1) (x + x + 1)
lim
= lim
·
x→1
x→1 (x − 1)
(x − 1)(x + 1)
(x + 1)
(x2 + x + 1)
3 3
(x − 1)
= [lim
] · [lim
]=1· =
x→1 (x − 1)
x→1
(x + 1)
2 2
Indeterminacy 0 · ∞:
lim[1 − cos x] · [cot x]
x→0
Indeterminacy 0 · ∞:
lim[1 − cos x] · [cot x]
x→0
factoring, re-grouping, and special limits:
cos x
1 − cos x x cos x
lim(1 − cos x)
= lim
x→0
sin x x→0
x
sin x 1
Indeterminacy 0 · ∞:
lim[1 − cos x] · [cot x]
x→0
factoring, re-grouping, and special limits:
cos x
1 − cos x x cos x
lim(1 − cos x)
= lim
x→0
sin x x→0
x
sin x 1
1 − cos x
x
cos x
= [lim
][lim
][lim
] = 0·1·1 = 0
x→0
x→0 sin x x→0 1
x
Indeterminacy ∞ − ∞:
p
p
lim [ x2 + x] − [ x2 − x]
x→∞
Indeterminacy ∞ − ∞:
p
p
lim [ x2 + x] − [ x2 − x]
x→∞
multiplying by conjugate, re-grouping:
√
√
√
√
( x2 + x − x2 − x) ( x2 + x + x2 − x)
√
√
= lim
x→∞
1
( x2 + x + x2 − x)
Indeterminacy ∞ − ∞:
p
p
lim [ x2 + x] − [ x2 − x]
x→∞
multiplying by conjugate, re-grouping:
√
√
√
√
( x2 + x − x2 − x) ( x2 + x + x2 − x)
√
√
= lim
x→∞
1
( x2 + x + x2 − x)
2
2
x +x−x +x
√
= lim √
x→∞ x2 + x +
x2 − x
2x
√
= lim √
x→∞ x2 + x +
x2 − x
2x
√
= lim √
x→∞ x2 + x +
x2 − x
∞/∞: division by the highest exponent
= lim r
x→∞
2x
xr
x2 x
+
+
x2 x2
x2 x
−
x2 x2
2
√
=√
=1
1+0+ 1−0
Infinite limits
infinite value of function
Definition. lim f (x) = ∞
x→c
⇔ ∀M > 0 ∃δ > 0 / 0 < |x − c| < δ ⇒ f (x) > M
Definition. lim f (x) = ∞
x→c
⇔ ∀M > 0 ∃δ > 0 / 0 < |x − c| < δ ⇒ f (x) > M
1
Example: lim
=∞
2
x→2 (x − 2)
⇔ ∀M > 0 ∃δ > 0 / 0 < |x−2| <
1
⇒
>
M
(x − 2)2
(in particular,
1
(x−2)2
> M if δ =
√1
M
)
Example 7. lim f (x) = −∞
x→c
⇔ ∀M < 0 ∃δ > 0 /
0 < |x − c| < δ ⇒ f (x) < M
Example 7. lim f (x) = −∞
x→c
⇔ ∀M < 0 ∃δ > 0 /
0 < |x − c| < δ ⇒ f (x) < M
Example 8. lim f (x) = ∞
x→∞
⇔ ∀M > 0 ∃N > 0 /
x > N ⇒ f (x) > M
EXAMPLES
EXAMPLE 9. Evaluate infinite limit
1
lim 2
x→1− x − 1
EXAMPLE 9. Evaluate infinite limit
1
lim 2
x→1− x − 1
Factoring and sign analysis:
1
1
= lim
= −
= −∞
(0 ) · (2)
x→1− (x − 1)(x + 1)
EXAMPLE 10. Evaluate infinite limit
x2 + 3x − 4
lim 2
x→2 x − 4x + 4
EXAMPLE 10. Evaluate infinite limit
x2 + 3x − 4
lim 2
x→2 x − 4x + 4
Factoring and sign analysis:
(x + 4)(x − 1) (6) · (1)
= lim
=
= −∞
2
+
x→2
(x − 2)
(0 )
EXAMPLE 11. Evaluate infinite limit
1
lim
x→0 sin x
EXAMPLE 11. Evaluate infinite limit
1
lim
x→0 sin x
Sign analysis for one-sided limits:
1
1
lim
= + = +∞
(0 )
x→0+ sin x
EXAMPLE 11. Evaluate infinite limit
1
lim
x→0 sin x
Sign analysis for one-sided limits:
1
1
lim
= + = +∞
(0 )
x→0+ sin x
1
1
lim
= − = −∞
(0 )
x→0− sin x
EXAMPLE 11. Evaluate infinite limit
1
lim
x→0 sin x
Sign analysis for one-sided limits:
1
1
lim
= + = +∞
(0 )
x→0+ sin x
1
1
lim
= − = −∞
(0 )
x→0− sin x
Limit at 0 does not exist