Lesson 4: Calculating Limits (handout)
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Transcript of Lesson 4: Calculating Limits (handout)
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Sec on 1.4Calcula ng Limits
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
February 2, 2011
Announcements
I First wri en HW due todayI Get-to-know-you survey and photo deadline is February 11
.
Announcements
I First wri en HW duetoday
I Get-to-know-you surveyand photo deadline isFebruary 11
.
ObjectivesI Know basic limits likelimx→a
x = a and limx→a
c = c.I Use the limit laws tocompute elementarylimits.
I Use algebra to simplifylimits.
I Understand and state theSqueeze Theorem.
I Use the SqueezeTheorem to demonstratea limit.
.
Notes
.
Notes
.
Notes
. 1.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
..
Limit
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OutlineRecall: The concept of limit
Basic Limits
Limit LawsThe direct subs tu on property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
.
Heuristic Definition of a 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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Notes
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Notes
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Notes
. 2.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
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The error-tolerance gameA game between two players (Dana and Emerson) to decide if a limitlimx→a
f(x) exists.
Step 1 Dana proposes L to be the limit.Step 2 Emerson challenges with an “error” level around L.Step 3 Dana chooses a “tolerance” level around a so that points x
within that tolerance of a (not coun ng a itself) are taken tovalues y within the error level of L. If Dana cannot, Emersonwins and the limit cannot be L.
Step 4 If Dana’s move is a good one, Emerson can challenge againor give up. If Emerson gives up, Dana wins and the limit is L.
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The error-tolerance game
..
This tolerance is too big
.
S ll too big
.
This looks good
.
So does this
.a
.
L
I To be legit, the part of the graph inside the blue (ver cal) stripmust also be inside the green (horizontal) strip.
I Even if Emerson shrinks the error, Dana can s ll move.
.
The error-tolerance game
.
.
This tolerance is too big
.
S ll too big
.
This looks good
.
So does this
.a
.
L
I To be legit, the part of the graph inside the blue (ver cal) stripmust also be inside the green (horizontal) strip.
I Even if Emerson shrinks the error, Dana can s ll move.
.
Notes
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Notes
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Notes
. 3.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
The error-tolerance game
.
.
This tolerance is too big
.
S ll too big
.
This looks good
.
So does this
.a
.
L
I To be legit, the part of the graph inside the blue (ver cal) stripmust also be inside the green (horizontal) strip.
I Even if Emerson shrinks the error, Dana can s ll move.
.
The error-tolerance game
.
.
This tolerance is too big
.
S ll too big
.
This looks good
.
So does this
.a
.
L
I To be legit, the part of the graph inside the blue (ver cal) stripmust also be inside the green (horizontal) strip.
I Even if Emerson shrinks the error, Dana can s ll move.
.
Limit FAIL: Jump
.. x.
y
..
−1
..
1
...
Part of graphinside blueis not insidegreen
.
Part of graphinside blueis not insidegreen
I So limx→0
|x|x
does notexist.
I But limx→0+
f(x) = 1
andlimx→0−
f(x) = −1.
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Notes
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Notes
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Notes
. 4.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Limit FAIL: Jump
.. x.
y
..
−1
..
1
..
.
Part of graphinside blueis not insidegreen
.
Part of graphinside blueis not insidegreen
I So limx→0
|x|x
does notexist.
I But limx→0+
f(x) = 1
andlimx→0−
f(x) = −1.
.
Limit FAIL: Jump
.. x.
y
..
−1
..
1
..
.
Part of graphinside blueis not insidegreen
.
Part of graphinside blueis not insidegreen
I So limx→0
|x|x
does notexist.
I But limx→0+
f(x) = 1
andlimx→0−
f(x) = −1.
.
Limit FAIL: unboundedness
.. x.
y
.0..
L?
.
The graph escapesthe green, so nogood
.
Even worse!
.
limx→0+
1xdoes not exist be-
cause the func on is un-bounded near 0
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Notes
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Notes
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Notes
. 5.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Limit EPIC FAILHere is a graph of the func on f(x) = sin
(πx
):
.. x.
y
..
−1
..
1
For every y in [−1, 1], there are infinitely many points x arbitrarilyclose to zero where f(x) = y. So lim
x→0f(x) cannot exist.
.
OutlineRecall: The concept of limit
Basic Limits
Limit LawsThe direct subs tu on property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
.
Really basic limits
FactLet c be a constant and a a real number.(i) lim
x→ax = a
(ii) limx→a
c = c
Proof.The first is tautological, the second is trivial.
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Notes
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Notes
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Notes
. 6.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
ET game for f(x) = x
.. x.
y
..a
..
a
I Se ng error equal to tolerance works!
.
ET game for f(x) = c
.. x.
y
..a
..
c
I any tolerance works!
.
OutlineRecall: The concept of limit
Basic Limits
Limit LawsThe direct subs tu on property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
.
Notes
.
Notes
.
Notes
. 7.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Limits and arithmetic
FactSuppose lim
x→af(x) = L and lim
x→ag(x) = M and c is a constant. Then
1. limx→a
[f(x) + g(x)] = L+M (errors add)
2. limx→a
[f(x)− g(x)] = L−M (combina on of adding and scaling)
3. limx→a
[cf(x)] = cL (error scales)
4. limx→a
[f(x)g(x)] = L ·M (more complicated, but doable)
.
Limits and arithmetic
FactSuppose lim
x→af(x) = L and lim
x→ag(x) = M and c is a constant. Then
1. limx→a
[f(x) + g(x)] = L+M (errors add)
2. limx→a
[f(x)− g(x)] = L−M (combina on of adding and scaling)
3. limx→a
[cf(x)] = cL (error scales)
4. limx→a
[f(x)g(x)] = L ·M (more complicated, but doable)
.
Justification of the scaling lawI errors scale: If f(x) is e away from L, then
(c · f(x)− c · L) = c · (f(x)− L) = c · e
That is, (c · f)(x) is c · e away from cL,I So if Emerson gives us an error of 1 (for instance), Dana can usethe fact that lim
x→af(x) = L to find a tolerance for f and g
corresponding to the error 1/c.I Dana wins the round.
.
Notes
.
Notes
.
Notes
. 8.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Limits and arithmetic
FactSuppose lim
x→af(x) = L and lim
x→ag(x) = M and c is a constant. Then
1. limx→a
[f(x) + g(x)] = L+M (errors add)
2. limx→a
[f(x)− g(x)] = L−M (combina on of adding and scaling)
3. limx→a
[cf(x)] = cL (error scales)
4. limx→a
[f(x)g(x)] = L ·M (more complicated, but doable)
.
Limits and arithmetic
FactSuppose lim
x→af(x) = L and lim
x→ag(x) = M and c is a constant. Then
1. limx→a
[f(x) + g(x)] = L+M (errors add)
2. limx→a
[f(x)− g(x)] = L−M (combina on of adding and scaling)
3. limx→a
[cf(x)] = cL (error scales)
4. limx→a
[f(x)g(x)] = L ·M (more complicated, but doable)
.
Limits and arithmetic IIFact (Con nued)
5. limx→a
f(x)g(x)
=LM
, if M ̸= 0.
6. limx→a
[f(x)]n =[limx→a
f(x)]n
(follows from 4 repeatedly)
7. limx→a
xn = an (follows from 6)
8. limx→a
n√x = n
√a
9. limx→a
n√
f(x) = n
√limx→a
f(x) (If n is even, we must addi onallyassume that lim
x→af(x) > 0)
.
Notes
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Notes
.
Notes
. 9.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Caution!I The quo ent rule for limits says that if lim
x→ag(x) ̸= 0, then
limx→a
f(x)g(x)
=limx→a f(x)limx→a g(x)
I It does NOT say that if limx→a
g(x) = 0, then
limx→a
f(x)g(x)
does not exist
In fact, limits of quo ents where numerator and denominatorboth tend to 0 are exactly where the magic happens.
I more about this later
.
Limits and arithmetic IIFact (Con nued)
5. limx→a
f(x)g(x)
=LM
, if M ̸= 0.
6. limx→a
[f(x)]n =[limx→a
f(x)]n
(follows from 4 repeatedly)
7. limx→a
xn = an (follows from 6)
8. limx→a
n√x = n
√a
9. limx→a
n√
f(x) = n
√limx→a
f(x) (If n is even, we must addi onallyassume that lim
x→af(x) > 0)
.
Applying the limit lawsExample
Find limx→3
(x2 + 2x+ 4
).
Solu onBy applying the limit laws repeatedly:
limx→3
(x2 + 2x+ 4
)= lim
x→3
(x2)+ lim
x→3(2x) + lim
x→3(4)
=(limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3+ 4= 9+ 6+ 4 = 19.
.
Notes
.
Notes
.
Notes
. 10.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Your turnExample
Find limx→3
x2 + 2x+ 4x3 + 11
Solu on
The answer is1938
=12.
.
Direct Substitution Property
As a direct consequence of the limit laws and the really basic limitswe have:Theorem (The Direct Subs tu on Property)
If f is a polynomial or a ra onal func on and a is in the domain of f,then
limx→a
f(x) = f(a)
.
OutlineRecall: The concept of limit
Basic Limits
Limit LawsThe direct subs tu on property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
.
Notes
.
Notes
.
Notes
. 11.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Limits do not see the point!
TheoremIf f(x) = g(x) when x ̸= a, and lim
x→ag(x) = L, then lim
x→af(x) = L.
.
Example of the MTP principleExample
Find limx→−1
x2 + 2x+ 1x+ 1
, if it exists.
Solu on
Sincex2 + 2x+ 1
x+ 1= x+ 1 whenever x ̸= −1, and since
limx→−1
x+ 1 = 0, we have limx→−1
x2 + 2x+ 1x+ 1
= 0.
.
ET game for f(x) =x2 + 2x + 1
x + 1
.. x.
y
...−1
I Even if f(−1) were something else, it would not effect the limit.
.
Notes
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Notes
.
Notes
. 12.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Limit of a function definedpiecewise at a boundary pointExample
Let
f(x) =
{x2 x ≥ 0−x x < 0
Does limx→0
f(x) exist?
..
Solu onWe have
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:limx→0−
f(x) = limx→0−
−x = −0 = 0
So limx→0
f(x) = 0.
.
Finding limits by algebraicmanipulations
Example
Find limx→4
√x− 2x− 4
.
Solu on
Write the denominator as x− 4 =√x2 − 4 = (
√x− 2)(
√x+ 2). So
limx→4
√x− 2x− 4
= limx→4
√x− 2
(√x− 2)(
√x+ 2)
= limx→4
1√x+ 2
=14
.
Your turnExample
Let
f(x) =
{1− x2 x ≥ 12x x < 1
Find limx→1
f(x) if it exists.
...1
..
Solu on
We have
limx→1+
f(x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
The le - and right-hand limits disagree, so the limit does not exist.
.
Notes
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Notes
.
Notes
. 13.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Two Important Limit TheoremsTheoremIf f(x) ≤ g(x) when x is near a (except possibly at a), then
limx→a
f(x) ≤ limx→a
g(x)
(as usual, provided these limits exist).
Theorem (The Squeeze/Sandwich/Pinching Theorem)
If f(x) ≤ g(x) ≤ h(x) when x is near a (as usual, except possibly ata), and
limx→a
f(x) = limx→a
h(x) = L,
thenlimx→a
g(x) = L.
.
Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicatedexpressions with simple ones when taking the limit.Example
Show that limx→0
x2 sin(πx
)= 0.
Solu onWe have for all x,
−1 ≤ sin(πx
)≤ 1 =⇒ −x2 ≤ x2 sin
(πx
)≤ x2
The le and right sides go to zero as x → 0.
.
Illustrating the Squeeze Theorem
.. x.
y
.
h(x) = x2
.
f(x) = −x2
.
g(x) = x2 sin(πx
)
.
Notes
.
Notes
.
Notes
. 14.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
OutlineRecall: The concept of limit
Basic Limits
Limit LawsThe direct subs tu on property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
.
Two important trigonometriclimits
TheoremThe following two limits hold:
I limθ→0
sin θθ
= 1
I limθ→0
cos θ − 1θ
= 0
.
Proof of the Sine LimitProof.
.. θ.sin θ
.cos θ
.θ
.tan θ
.−1
.1
No ce
sin θ ≤ θ ≤ 2 tanθ
2≤ tan θ
Divide by sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Take reciprocals:
1 ≥ sin θθ
≥ cos θ
As θ → 0, the le and right sides tend to 1. So, then, must themiddle expression.
.
Notes
.
Notes
.
Notes
. 15.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011
.
.
Proof of the Cosine LimitProof.
1− cos θθ
=1− cos θ
θ· 1+ cos θ1+ cos θ
=1− cos2 θθ(1+ cos θ)
=sin2 θ
θ(1+ cos θ)=
sin θθ
· sin θ1+ cos θ
So
limθ→0
1− cos θθ
=
(limθ→0
sin θθ
)·(limθ→0
sin θ1+ cos θ
)= 1 · 0 = 0.
.
Try theseExample
1. limθ→0
tan θθ
2. limθ→0
sin 2θθ
Answer
1. 12. 2
.
Summary
I The limit laws allow us tocompute limitsreasonably.
I BUT we cannot make upextra laws otherwise weget into trouble.
.. x.
y
.
Notes
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Notes
.
Notes
. 16.
. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011