Limits at Infinity and Infinite Limits - csun.eduama5348/calculus/presentations/math... · Limits...
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Limits at Infinity and Infinite Limits more examples of limits
Transcript of Limits at Infinity and Infinite Limits - csun.eduama5348/calculus/presentations/math... · Limits...
Limits at Infinity
and Infinite Limits
more examples of limits
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Motivation:
handling infinite variable
and infinite function
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Question. Can we describein mathematics:
(1) infinite value of variable?
(2) infinite value of function?
O
f(x)=1/x
1
2
Question. Can we describein mathematics:
(1) infinite value of variable?
(2) infinite value of function?
O
f(x)=1/x
1
2
Application: horizontal and vertical asymptotes
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Limits at infinity
infinite value of variable
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Definition. L = limx→∞
f(x)
⇔ ∀ε > 0 ∃N > 0 / x > N ⇒ |f(x)− L| < ε
Definition. L = limx→∞
f(x)
⇔ ∀ε > 0 ∃N > 0 / x > N ⇒ |f(x)− L| < ε
Example: limx→∞
5x + 1x− 2
= 5
⇔ ∀ε > 0 ∃N > 0 / x > N
⇒∣∣∣∣5x + 1x− 2
− 5∣∣∣∣ < ε
(in particular, |5x+1x−2 − 5| = 11
|x−2| < ε if N = 2 + 11ε )
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Properties: Let limx→∞
f(x) and limx→∞
g(x) exist,
limx→∞
[f(x) + g(x)] = limx→∞
f(x) + limx→∞
g(x)
limx→∞
[f(x)− g(x)] = limx→∞
f(x)− limx→∞
g(x)
limx→∞
[f(x) · g(x)] = [ limx→∞
f(x)] · [ limx→∞
g(x)]
limx→∞
[f(x)/g(x)] = [ limx→∞
f(x)]/[ limx→∞
g(x)],limx→∞
g(x) 6= 0
limx→∞
n√
g(x) = n
√limx→∞
g(x), limx→∞
g(x) > 0, n even
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EXAMPLES
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EXAMPLE 1. Evaluate limit
limx→∞
1x
EXAMPLE 1. Evaluate limit
limx→∞
1x
As variable x gets larger, 1/x gets smaller because1 is being divided by a laaaaaaaarge number:(
x = 1010,1x
=1
1010
)
EXAMPLE 1. Evaluate limit
limx→∞
1x
As variable x gets larger, 1/x gets smaller because1 is being divided by a laaaaaaaarge number:(
x = 1010,1x
=1
1010
)The limit is 0.
limx→∞
1x
= 0.
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EXAMPLE 2. Limits
limx→∞
1xn
= 0
limx→∞
1n√
x= 0
EXAMPLE 2. Limits
limx→∞
1xn
= 0
limx→∞
1n√
x= 0
By the same argument
limx→∞
1x− 100
= 0, limx→∞
1n√
x− 10000= 0
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EXAMPLE 3. Evaluate limit
limx→∞
1n√
x2 + x− C
EXAMPLE 3. Evaluate limit
limx→∞
1n√
x2 + x− C
Recall the graph of
y = x2 + x− C
Observation:
x →∞⇒ y →∞
Value of C does not matter.The answer is 0.
O
y= x + x - C2
-1/4-C
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EXAMPLE 4. Evaluate limit (not evoke graphs)
limx→∞
1x2 − x− 1
EXAMPLE 4. Evaluate limit (not evoke graphs)
limx→∞
1x2 − x− 1
Attention: indeterminacy ∞−∞
= limx→∞
1
x2(1− 1
x− 1
x2
)= lim
x→∞
1x2 lim
x→∞
1
1− 1x− 1
x2
= 0 · 1 = 0
– Typeset by FoilTEX – 11
EXAMPLE 5. Evaluate limit
limx→∞
5x3 − 2x2 − 1x3 − x + 1
EXAMPLE 5. Evaluate limit
limx→∞
5x3 − 2x2 − 1x3 − x + 1
Standard trick: divide by the highest degree of x
= limx→∞
5x3
x3 −2x2
x3 −1x3
x3
x3 −x
x3 +1x3
= limx→∞
5− 2x− 1
x3
1− 1x2 +
1x3
=5− 0− 01− 0 + 0
= 5
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EXAMPLE 6. Evaluate limit
limx→∞
x− 2√2x2 − x + 1
EXAMPLE 6. Evaluate limit
limx→∞
x− 2√2x2 − x + 1
Standard trick: divide by the highest degree of x
= limx→∞
x
x− 2
x√2x2 − x + 1√
x2
= limx→∞
x
x− 2
x3√2x2
x2 −x
x2 +1x2
=1− 0√
2− 0 + 0=
1√2.
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Indeterminacy:
unsuitable breaking in limits
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Attention: Indeterminacy
limx→∞
[x2 + 1][x3 − 3]
( ∞∞
)limx→1
[x3 − 1][x2 − 1]
( 00
)Wrong breaking! Limit laws do not apply!
limx→0
[1− cos x][cot x] ( 0 · ∞ )
Substitution is undefined!
limx→∞
[√
x2 + x]− [√
x2 − x] ( ∞−∞ )
Rewrite first!
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Indeterminacy∞∞
:
limx→∞
[x2 + 1][x3 − 3]
Indeterminacy∞∞
:
limx→∞
[x2 + 1][x3 − 3]
division by highest exponent of x:
limx→∞
x2
x3 +1x3
x3
x3 −3x3
=0 + 01− 0
= 0
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Indeterminacy00:
limx→1
[x3 − 1][x2 − 1]
Indeterminacy00:
limx→1
[x3 − 1][x2 − 1]
factoring and re-grouping:
limx→1
(x− 1)(x2 + x + 1)(x− 1)(x + 1)
= limx→1
(x− 1)(x− 1)
·(x2 + x + 1)(x + 1)
= [limx→1
(x− 1)(x− 1)
] · [limx→1
(x2 + x + 1)(x + 1)
] = 1 · 32
=32
– Typeset by FoilTEX – 17
Indeterminacy 0 · ∞:
limx→0
[1− cos x] · [cot x]
Indeterminacy 0 · ∞:
limx→0
[1− cos x] · [cot x]
factoring, re-grouping, and special limits:
limx→0
(1− cos x)cos x
sinx= lim
x→0
1− cos x
x
x
sinx
cos x
1
Indeterminacy 0 · ∞:
limx→0
[1− cos x] · [cot x]
factoring, re-grouping, and special limits:
limx→0
(1− cos x)cos x
sinx= lim
x→0
1− cos x
x
x
sinx
cos x
1
= [limx→0
1− cos x
x][lim
x→0
x
sinx][lim
x→0
cos x
1] = 0 ·1 ·1 = 0
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Indeterminacy ∞−∞:
limx→∞
[√
x2 + x]− [√
x2 − x]
Indeterminacy ∞−∞:
limx→∞
[√
x2 + x]− [√
x2 − x]
multiplying by conjugate, re-grouping:
= limx→∞
(√
x2 + x−√
x2 − x)1
(√
x2 + x +√
x2 − x)(√
x2 + x +√
x2 − x)
Indeterminacy ∞−∞:
limx→∞
[√
x2 + x]− [√
x2 − x]
multiplying by conjugate, re-grouping:
= limx→∞
(√
x2 + x−√
x2 − x)1
(√
x2 + x +√
x2 − x)(√
x2 + x +√
x2 − x)
= limx→∞
x2 + x− x2 + x√x2 + x +
√x2 − x
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= limx→∞
2x√x2 + x +
√x2 − x
= limx→∞
2x√x2 + x +
√x2 − x
∞/∞: division by the highest exponent
= limx→∞
2x
x√x2
x2 +x
x2 +
√x2
x2 −x
x2
=2√
1 + 0 +√
1− 0= 1
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Infinite limits
infinite value of function
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Definition. limx→c
f(x) = ∞
⇔ ∀M > 0 ∃δ > 0 / 0 < |x− c| < δ ⇒ f(x) > M
Definition. limx→c
f(x) = ∞
⇔ ∀M > 0 ∃δ > 0 / 0 < |x− c| < δ ⇒ f(x) > M
Example: limx→2
1(x− 2)2 = ∞
⇔ ∀M > 0 ∃δ > 0 / 0 < |x−2| < δ
⇒ 1(x− 2)2 > M
(in particular, 1(x−2)2 > M if δ = 1√
M)
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Example 7. limx→c
f(x) = −∞
⇔ ∀M < 0 ∃δ > 0 /
0 < |x− c| < δ ⇒ f(x) < M
Example 7. limx→c
f(x) = −∞
⇔ ∀M < 0 ∃δ > 0 /
0 < |x− c| < δ ⇒ f(x) < M
Example 8. limx→∞
f(x) = ∞
⇔ ∀M > 0 ∃N > 0 /
x > N ⇒ f(x) > M
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EXAMPLES
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EXAMPLE 9. Evaluate infinite limit
limx→1−
1x2 − 1
EXAMPLE 9. Evaluate infinite limit
limx→1−
1x2 − 1
Factoring and sign analysis:
= limx→1−
1(x− 1)(x + 1)
=1
(0−) · (2)= −∞
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EXAMPLE 10. Evaluate infinite limit
limx→2
x2 + 3x− 4x2 − 4x + 4
EXAMPLE 10. Evaluate infinite limit
limx→2
x2 + 3x− 4x2 − 4x + 4
Factoring and sign analysis:
= limx→2
(x + 4)(x− 1)(x− 2)2 =
(6) · (1)(0+)
= −∞
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EXAMPLE 11. Evaluate infinite limit
limx→0
1sinx
EXAMPLE 11. Evaluate infinite limit
limx→0
1sinx
Sign analysis for one-sided limits:
limx→0+
1sinx
=1
(0+)= +∞
EXAMPLE 11. Evaluate infinite limit
limx→0
1sinx
Sign analysis for one-sided limits:
limx→0+
1sinx
=1
(0+)= +∞
limx→0−
1sinx
=1
(0−)= −∞
EXAMPLE 11. Evaluate infinite limit
limx→0
1sinx
Sign analysis for one-sided limits:
limx→0+
1sinx
=1
(0+)= +∞
limx→0−
1sinx
=1
(0−)= −∞
Limit at 0 does not exist
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