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Sec on 1.5Limits Involving Infinity
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
February 9, 2011
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Announcements
I Get-to-know-you extracredit due FridayFebruary 11
I Quiz 1 is next week inrecita on. CoversSec ons 1.1–1.4
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Objectives
I “Intuit” limits involving infinity byeyeballing the expression.
I Show limits involving infinity byalgebraic manipula on and conceptualargument.
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Notes
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Notes
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Notes
. 1.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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Recall the definition of limitDefini onWe write
limx→a
f(x) = L
and say
“the limit of f(x), as x approaches a, equals L”
if we can make the values of f(x) arbitrarily close to L (as close to Las we like) by taking x to be sufficiently close to a (on either side ofa) but not equal to a.
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The unboundedness problem
Recall why limx→0+
1xdoesn’t
exist.No ma er how thin we drawthe strip to the right of x = 0,we cannot “capture” the graphinside the box.
.. x.
y
..
L?
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Infinite LimitsDefini onThe nota on
limx→a
f(x) = ∞
means that values of f(x) can bemade arbitrarily large (as large as weplease) by taking x sufficiently closeto a but not equal to a. .. x.
y
I “Large” takes the place of “close to L”.
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Notes
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Notes
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Notes
. 2.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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Negative InfinityDefini onThe nota on
limx→a
f(x) = −∞
means that the values of f(x) can be made arbitrarily large nega ve(as large as we please) by taking x sufficiently close to a but notequal to a.
I We call a number large or small based on its absolute value. So−1, 000, 000 is a large (nega ve) number.
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Vertical Asymptotes
Defini onThe line x = a is called a ver cal asymptote of the curve y = f(x) ifat least one of the following is true:
I limx→a
f(x) = ∞I lim
x→a+f(x) = ∞
I limx→a−
f(x) = ∞
I limx→a
f(x) = −∞I lim
x→a+f(x) = −∞
I limx→a−
f(x) = −∞
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Infinite Limits we Know
I limx→0+
1x= ∞
I limx→0−
1x= −∞
I limx→0
1x2
= ∞
.. x.
y
............
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Notes
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Notes
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Notes
. 3.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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Finding limits at trouble spotsExample
Letf(x) =
x2 + 2x2 − 3x+ 2
Find limx→a−
f(x) and limx→a+
f(x) for each a at which f is not con nuous.
Solu onThe denominator factors as (x− 1)(x− 2). We can record the signsof the factors on the number line.
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Use the number line
.. (x− 1).−. small
. small
..1
. 0. +.
(x− 2)
.−..
2
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0
.
small.
small.
+
.
(x2 + 2)
.
+
.
f(x)
..
1
..
2
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+
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+∞
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−∞
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−
.
−∞
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+∞
.
+
limx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
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In English, now
To explain the limit, you can say:“As x → 1−, the numerator approaches 3, and the denominatorapproaches 0 while remaining posi ve. So the limit is +∞.”
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Notes
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Notes
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Notes
. 4.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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The graph so farlimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
.. x.
y
..−1
..1
..2
..3
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Rules of Thumb with infinite limitsFactThe sum of two posi ve or two nega ve infinite limits is infinite.
I If limx→a
f(x) = ∞ and limx→a
g(x) = ∞, then limx→a
(f(x) + g(x)) = ∞.
..
∞+∞ = ∞
I If limx→a
f(x) = −∞ and limx→a
g(x) = −∞, thenlimx→a
(f(x) + g(x)) = −∞.
..
−∞+ (−∞) = −∞
RemarkWe don’t say anything here about limits of the form∞−∞.
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Rules of Thumb with infinite limits
FactThe sum of a finite limit and an infinite limit is infinite.
I If limx→a
f(x) = L and limx→a
g(x) = ±∞,
..
L+∞ = ∞L−∞ = −∞
thenlimx→a
(f(x) + g(x)) = ±∞.
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Notes
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Notes
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Notes
. 5.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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Rules of Thumb with infinite limitsFactThe product of a finite limit and an infinite limit is infinite if the finitelimit is not 0.
...
L · ∞ =
{∞ if L > 0−∞ if L < 0.
I If limx→a
f(x) = L, limx→a
g(x) = ∞, and L > 0, thenlimx→a
f(x) · g(x) = ∞.
I If limx→a
f(x) = L, limx→a
g(x) = ∞, and L < 0, thenlimx→a
f(x) · g(x) = −∞.
I If limx→a
f(x) = L, limx→a
g(x) = −∞, and L > 0, thenlimx→a
f(x) · g(x) = −∞.
I If limx→a
f(x) = L, limx→a
g(x) = −∞, and L < 0, thenlimx→a
f(x) · g(x) = ∞.
..
L · (−∞) =
{−∞ if L > 0∞ if L < 0.
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Multiplying infinite limits
FactThe product of two infinite limits is infinite.
..
∞ ·∞ = ∞∞ · (−∞) = −∞
(−∞) · (−∞) = ∞
I If limx→a
f(x) = ∞ and limx→a
g(x) = ∞, then limx→a
f(x) · g(x) = ∞.
I If limx→a
f(x) = ∞ and limx→a
g(x) = −∞, then limx→a
f(x) · g(x) = −∞.
I If limx→a
f(x) = −∞ and limx→a
g(x) = −∞, then limx→a
f(x) · g(x) = ∞.
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Dividing by Infinity
FactThe quo ent of a finite limit by an infinite limit is zero.
I If limx→a
f(x) = L and limx→a
g(x) = ±∞, then limx→a
f(x)g(x)
= 0.
..
L∞
= 0
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Notes
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Notes
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Notes
. 6.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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Dividing by zero is still not allowed
..10 =∞
There are examples of such limit forms where the limit is∞,−∞,undecided between the two, or truly neither.
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Indeterminate Limit formsLimits of the form
L0are indeterminate. There is no rule for
evalua ng such a form; the limit must be examined more closely.Consider these:
limx→0
1x2
= ∞ limx→0
−1x2
= −∞
limx→0+
1x= ∞ lim
x→0−
1x= −∞
Worst, limx→0
1x sin(1/x)
is of the formL0, but the limit does not exist,
even in the le - or right-hand sense. There are infinitely manyver cal asymptotes arbitrarily close to 0!
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Indeterminate Limit formsLimits of the form 0 · ∞ and∞−∞ are also indeterminate.
Example
I The limit limx→0+
sin x · 1xis of the form 0 · ∞, but the answer is 1.
I The limit limx→0+
sin2 x · 1xis of the form 0 · ∞, but the answer is 0.
I The limit limx→0+
sin x · 1x2
is of the form 0 ·∞, but the answer is∞.
Limits of indeterminate forms may or may not “exist.” It will dependon the context.
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Notes
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Notes
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Notes
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. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
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OutlineVer cal AsymptotesInfinite Limits we KnowLimit “Laws” with Infinite LimitsIndeterminate Limit forms
Limits at∞Algebraic rates of growthRa onalizing to get a limit
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Limits at Infinity
Defini onLet f be a func on defined on some interval (a,∞). Then
limx→∞
f(x) = L
means that the values of f(x) can be made as close to L as we like, bytaking x sufficiently large.
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Notes
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Notes
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Notes
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. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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Horizontal Asymptotes
Defini onThe line y = L is a called a horizontal asymptote of the curvey = f(x) if either
limx→∞
f(x) = L or limx→−∞
f(x) = L.
y = L is a horizontal line!
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Basic limits at infinity
TheoremLet n be a posi ve integer. Then
I limx→∞
1xn
= 0
I limx→−∞
1xn
= 0
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Limit laws at infinityFactAny limit law that concerns finite limits at a finite point a is s ll trueif the finite point is replaced by±∞.That is, if lim
x→∞f(x) = L and lim
x→∞g(x) = M, then
I limx→∞
(f(x) + g(x)) = L+M
I limx→∞
(f(x)− g(x)) = L−M
I limx→∞
cf(x) = c · L (for any constant c)I lim
x→∞f(x) · g(x) = L ·M
I limx→∞
f(x)g(x)
=LM
(if M ̸= 0), etc.
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Notes
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Notes
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Notes
. 9.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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Using the limit laws to computelimits at ∞
Example
Find limx→∞
xx2 + 1
AnswerThe limit is 0. No ce thatthe graph does cross theasymptote, whichcontradicts one of thecommonly held beliefs ofwhat an asymptote is.
.. x.
y
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SolutionSolu on
Factor out the largest power of x from the numerator anddenominator. We have
xx2 + 1
=x(1)
x2(1+ 1/x2)=
1x· 11+ 1/x2
limx→∞
xx2 + 1
= limx→∞
1x
11+ 1/x2
= limx→∞
1x· limx→∞
11+ 1/x2
= 0 · 11+ 0
= 0.
RemarkHad the higher power been in the numerator, the limit would havebeen∞.
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Another ExampleExample
Findlimx→∞
2x3 + 3x+ 14x3 + 5x2 + 7
if it exists.A does not existB 1/2
C 0D ∞
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Notes
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Notes
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Notes
. 10.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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SolutionSolu on
Factor out the largest power of x from the numerator anddenominator. We have
2x3 + 3x+ 14x3 + 5x2 + 7
=x3(2+ 3/x2 + 1/x3)
x3(4+ 5/x + 7/x3)
limx→∞
2x3 + 3x+ 14x3 + 5x2 + 7
= limx→∞
2+ 3/x2 + 1/x3
4+ 5/x + 7/x3
=2+ 0+ 04+ 0+ 0
=12
Upshot
When finding limits of algebraic expressions at infinity, look at thehighest degree terms.
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Still Another Example
Example
Find
limx→∞
√3x4 + 7x2 + 3
..
√3x4 + 7 ∼
√3x4 =
√3x2
AnswerThe limit is
√3.
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SolutionSolu on
limx→∞
√3x4 + 7x2 + 3
= limx→∞
√x4(3+ 7/x4)
x2(1+ 3/x2)
= limx→∞
x2√
(3+ 7/x4)
x2(1+ 3/x2)
= limx→∞
√(3+ 7/x4)
1+ 3/x2
=
√3+ 01+ 0
=√3.
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Notes
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Notes
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Notes
. 11.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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Rationalizing to get a limit..Example
Compute limx→∞
(√4x2 + 17− 2x
).
Solu onThis limit is of the form∞−∞, which we cannot use. So we ra onalize thenumerator (the denominator is 1) to get an expression that we can use thelimit laws on.
limx→∞
(√4x2 + 17− 2x
)= lim
x→∞
(√4x2 + 17− 2x
)·√4x2 + 17+ 2x√4x2 + 17+ 2x
= limx→∞
(4x2 + 17)− 4x2√4x2 + 17+ 2x
= limx→∞
17√4x2 + 17+ 2x
= 0
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Kick it up a notchExample
Compute limx→∞
(√4x2 + 17x− 2x
).
Solu on
Same trick, different answer:
limx→∞
(√4x2 + 17x− 2x
)= lim
x→∞
(√4x2 + 17x− 2x
)·√4x2 + 17+ 2x√4x2 + 17x+ 2x
= limx→∞
(4x2 + 17x)− 4x2√4x2 + 17x+ 2x
= limx→∞
17x√4x2 + 17x+ 2x
= limx→∞
17√4+ 17/x+ 2
=174
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Summary
I Infinity is a more complicated concept than a single number.There are rules of thumb, but there are also excep ons.
I Take a two-pronged approach to limits involving infinity:I Look at the expression to guess the limit.I Use limit rules and algebra to verify it.
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Notes
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Notes
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Notes
. 12.
. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011
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