Download - Lesson 3: Limit Laws
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Section 1.4Calculating Limits
V63.0121.002.2010Su, Calculus I
New York University
May 18, 2010
Announcements
I WebAssign Class Key: nyu 0127 7953I Office Hours: MR 5:00–5:45, TW 7:50–8:30, CIWW 102 (here)I Quiz 1 Thursday on 1.1–1.4
. . . . . .
![Page 2: Lesson 3: Limit Laws](https://reader033.vdocument.in/reader033/viewer/2022052619/5562d812d8b42a49398b51b4/html5/thumbnails/2.jpg)
. . . . . .
Announcements
I WebAssign Class Key: nyu0127 7953
I Office Hours: MR5:00–5:45, TW 7:50–8:30,CIWW 102 (here)
I Quiz 1 Thursday on1.1–1.4
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 2 / 37
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. . . . . .
Objectives
I Know basic limits likelimx→a
x = a and limx→a
c = c.
I Use the limit laws tocompute elementary limits.
I Use algebra to simplifylimits.
I Understand and state theSqueeze Theorem.
I Use the Squeeze Theoremto demonstrate a limit.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 3 / 37
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. . . . . .
Outline
Basic Limits
Limit LawsThe direct substitution property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 4 / 37
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. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 5 / 37
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. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 5 / 37
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. . . . . .
ET game for f(x) = x
. .x
.y
..a
..a
I Setting error equal to tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 6 / 37
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. . . . . .
ET game for f(x) = x
. .x
.y
..a
..a
I Setting error equal to tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 6 / 37
![Page 9: Lesson 3: Limit Laws](https://reader033.vdocument.in/reader033/viewer/2022052619/5562d812d8b42a49398b51b4/html5/thumbnails/9.jpg)
. . . . . .
ET game for f(x) = x
. .x
.y
..a
..a
I Setting error equal to tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 6 / 37
![Page 10: Lesson 3: Limit Laws](https://reader033.vdocument.in/reader033/viewer/2022052619/5562d812d8b42a49398b51b4/html5/thumbnails/10.jpg)
. . . . . .
ET game for f(x) = x
. .x
.y
..a
..a
I Setting error equal to tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 6 / 37
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. . . . . .
ET game for f(x) = x
. .x
.y
..a
..a
I Setting error equal to tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 6 / 37
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. . . . . .
ET game for f(x) = x
. .x
.y
..a
..a
I Setting error equal to tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 6 / 37
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. . . . . .
ET game for f(x) = x
. .x
.y
..a
..a
I Setting error equal to tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 6 / 37
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. . . . . .
ET game for f(x) = c
.
.x
.y
..a
..c
I any tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 7 / 37
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. . . . . .
ET game for f(x) = c
. .x
.y
..a
..c
I any tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 7 / 37
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. . . . . .
ET game for f(x) = c
. .x
.y
..a
..c
I any tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 7 / 37
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. . . . . .
ET game for f(x) = c
. .x
.y
..a
..c
I any tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 7 / 37
![Page 18: Lesson 3: Limit Laws](https://reader033.vdocument.in/reader033/viewer/2022052619/5562d812d8b42a49398b51b4/html5/thumbnails/18.jpg)
. . . . . .
ET game for f(x) = c
. .x
.y
..a
..c
I any tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 7 / 37
![Page 19: Lesson 3: Limit Laws](https://reader033.vdocument.in/reader033/viewer/2022052619/5562d812d8b42a49398b51b4/html5/thumbnails/19.jpg)
. . . . . .
ET game for f(x) = c
. .x
.y
..a
..c
I any tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 7 / 37
![Page 20: Lesson 3: Limit Laws](https://reader033.vdocument.in/reader033/viewer/2022052619/5562d812d8b42a49398b51b4/html5/thumbnails/20.jpg)
. . . . . .
ET game for f(x) = c
. .x
.y
..a
..c
I any tolerance works!
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 7 / 37
![Page 21: Lesson 3: Limit Laws](https://reader033.vdocument.in/reader033/viewer/2022052619/5562d812d8b42a49398b51b4/html5/thumbnails/21.jpg)
. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 8 / 37
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. . . . . .
Outline
Basic Limits
Limit LawsThe direct substitution property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 9 / 37
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. . . . . .
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
(combination 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 10 / 37
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. . . . . .
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
(combination 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 10 / 37
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. . . . . .
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
(combination 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 11 / 37
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. . . . . .
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
(combination 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 11 / 37
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. . . . . .
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
(combination 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 11 / 37
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. . . . . .
Justification of the scaling law
I 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 Player 2 gives us an error of 1 (for instance), Player 1 can
use the fact that limx→a
f(x) = L to find a tolerance for f and gcorresponding to the error 1/c.
I Player 1 wins the round.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 12 / 37
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. . . . . .
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 (combination 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 13 / 37
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. . . . . .
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 (combination 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 14 / 37
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. . . . . .
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 (combination 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 14 / 37
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. . . . . .
Limits and arithmetic II
Fact (Continued)
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 additionally assumethat lim
x→af(x) > 0)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 15 / 37
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. . . . . .
Caution!
I The quotient rule for limits says that if limx→a
g(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, it can.I more about this later
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 16 / 37
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. . . . . .
Limits and arithmetic II
Fact (Continued)
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 additionally assumethat lim
x→af(x) > 0)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
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. . . . . .
Limits and arithmetic II
Fact (Continued)
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 additionally assumethat lim
x→af(x) > 0)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
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. . . . . .
Limits and arithmetic II
Fact (Continued)
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 additionally assumethat lim
x→af(x) > 0)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
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. . . . . .
Limits and arithmetic II
Fact (Continued)
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 additionally assumethat lim
x→af(x) > 0)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
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. . . . . .
Limits and arithmetic II
Fact (Continued)
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 additionally assumethat lim
x→af(x) > 0)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
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. . . . . .
Limits and arithmetic II
Fact (Continued)
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 additionally assumethat lim
x→af(x) > 0)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 17 / 37
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. . . . . .
Applying the limit laws
Example
Find limx→3
(x2 + 2x+ 4
).
SolutionBy applying the limit laws repeatedly:
limx→3
(x2 + 2x+ 4
)
= limx→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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
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. . . . . .
Applying the limit laws
Example
Find limx→3
(x2 + 2x+ 4
).
SolutionBy applying the limit laws repeatedly:
limx→3
(x2 + 2x+ 4
)
= limx→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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
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. . . . . .
Applying the limit laws
Example
Find limx→3
(x2 + 2x+ 4
).
SolutionBy 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
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. . . . . .
Applying the limit laws
Example
Find limx→3
(x2 + 2x+ 4
).
SolutionBy 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
![Page 44: Lesson 3: Limit Laws](https://reader033.vdocument.in/reader033/viewer/2022052619/5562d812d8b42a49398b51b4/html5/thumbnails/44.jpg)
. . . . . .
Applying the limit laws
Example
Find limx→3
(x2 + 2x+ 4
).
SolutionBy 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
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. . . . . .
Applying the limit laws
Example
Find limx→3
(x2 + 2x+ 4
).
SolutionBy 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 18 / 37
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. . . . . .
Your turn
Example
Find limx→3
x2 + 2x+ 4x3 + 11
Solution
The answer is1938
=12.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 19 / 37
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. . . . . .
Your turn
Example
Find limx→3
x2 + 2x+ 4x3 + 11
Solution
The answer is1938
=12.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 19 / 37
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. . . . . .
Direct Substitution Property
Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of f, then
limx→a
f(x) = f(a)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 20 / 37
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. . . . . .
Outline
Basic Limits
Limit LawsThe direct substitution property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 21 / 37
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. . . . . .
Limits do not see the point! (in a good way)
TheoremIf f(x) = g(x) when x ̸= a, and lim
x→ag(x) = L, then lim
x→af(x) = L.
Example
Find limx→−1
x2 + 2x+ 1x+ 1
, if it exists.
Solution
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 22 / 37
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. . . . . .
Limits do not see the point! (in a good way)
TheoremIf f(x) = g(x) when x ̸= a, and lim
x→ag(x) = L, then lim
x→af(x) = L.
Example
Find limx→−1
x2 + 2x+ 1x+ 1
, if it exists.
Solution
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 22 / 37
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. . . . . .
Limits do not see the point! (in a good way)
TheoremIf f(x) = g(x) when x ̸= a, and lim
x→ag(x) = L, then lim
x→af(x) = L.
Example
Find limx→−1
x2 + 2x+ 1x+ 1
, if it exists.
Solution
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 22 / 37
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. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 23 / 37
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. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 23 / 37
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. . . . . .
Limit of a function defined piecewise at a boundary
point
Example
Let
f(x) =
{x2 x ≥ 0−x x < 0
Does limx→0
f(x) exist?
.
.
SolutionWe have
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:lim
x→0−f(x) = lim
x→0−−x = −0 = 0
So limx→0
f(x) = 0.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
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. . . . . .
Limit of a function defined piecewise at a boundary
point
Example
Let
f(x) =
{x2 x ≥ 0−x x < 0
Does limx→0
f(x) exist?
.
.
SolutionWe have
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:lim
x→0−f(x) = lim
x→0−−x = −0 = 0
So limx→0
f(x) = 0.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
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. . . . . .
Limit of a function defined piecewise at a boundary
point
Example
Let
f(x) =
{x2 x ≥ 0−x x < 0
Does limx→0
f(x) exist?
.
.
SolutionWe have
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:lim
x→0−f(x) = lim
x→0−−x = −0 = 0
So limx→0
f(x) = 0.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
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. . . . . .
Limit of a function defined piecewise at a boundary
point
Example
Let
f(x) =
{x2 x ≥ 0−x x < 0
Does limx→0
f(x) exist?
..
SolutionWe have
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:lim
x→0−f(x) = lim
x→0−−x = −0 = 0
So limx→0
f(x) = 0.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
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. . . . . .
Limit of a function defined piecewise at a boundary
point
Example
Let
f(x) =
{x2 x ≥ 0−x x < 0
Does limx→0
f(x) exist?
..
SolutionWe have
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:lim
x→0−f(x) = lim
x→0−−x = −0 = 0
So limx→0
f(x) = 0.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
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. . . . . .
Limit of a function defined piecewise at a boundary
point
Example
Let
f(x) =
{x2 x ≥ 0−x x < 0
Does limx→0
f(x) exist?
.
.
SolutionWe have
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:lim
x→0−f(x) = lim
x→0−−x = −0 = 0
So limx→0
f(x) = 0.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
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. . . . . .
Limit of a function defined piecewise at a boundary
point
Example
Let
f(x) =
{x2 x ≥ 0−x x < 0
Does limx→0
f(x) exist?
.
.
SolutionWe have
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:lim
x→0−f(x) = lim
x→0−−x = −0 = 0
So limx→0
f(x) = 0.V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37
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. . . . . .
Finding limits by algebraic manipulations
Example
Find limx→4
√x− 2x− 4
.
SolutionWrite 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
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 25 / 37
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. . . . . .
Finding limits by algebraic manipulations
Example
Find limx→4
√x− 2x− 4
.
SolutionWrite 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
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 25 / 37
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. . . . . .
Finding limits by algebraic manipulations
Example
Find limx→4
√x− 2x− 4
.
SolutionWrite 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
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 25 / 37
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. . . . . .
Your turn
Example
Let
f(x) =
{1− x2 x ≥ 12x x < 1
Find limx→1
f(x) if it exists.
.
..1
.
.
SolutionWe have
limx→1+
f(x) = limx→1+
(1− x2
)DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
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. . . . . .
Your turn
Example
Let
f(x) =
{1− x2 x ≥ 12x x < 1
Find limx→1
f(x) if it exists.
.
..1
.
.
SolutionWe have
limx→1+
f(x) = limx→1+
(1− x2
)DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
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. . . . . .
Your turn
Example
Let
f(x) =
{1− x2 x ≥ 12x x < 1
Find limx→1
f(x) if it exists.
. ..1
.
.
SolutionWe have
limx→1+
f(x) = limx→1+
(1− x2
)DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
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. . . . . .
Your turn
Example
Let
f(x) =
{1− x2 x ≥ 12x x < 1
Find limx→1
f(x) if it exists.
. ..1
.
.
SolutionWe have
limx→1+
f(x) = limx→1+
(1− x2
)DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
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. . . . . .
Your turn
Example
Let
f(x) =
{1− x2 x ≥ 12x x < 1
Find limx→1
f(x) if it exists.
. ..1
.
.
SolutionWe have
limx→1+
f(x) = limx→1+
(1− x2
)DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
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. . . . . .
Your turn
Example
Let
f(x) =
{1− x2 x ≥ 12x x < 1
Find limx→1
f(x) if it exists.
. ..1
.
.
SolutionWe have
limx→1+
f(x) = limx→1+
(1− x2
)DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
The left- and right-hand limits disagree, so the limit does not exist.V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37
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. . . . . .
A message from the Mathematical Grammar Police
Please do not say “ limx→a
f(x) = DNE.” Does not compute.
I Too many verbsI Leads to FALSE limit laws like “If lim
x→af(x) DNE and lim
x→ag(x) DNE,
then limx→a
(f(x) + g(x)) DNE.”
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 27 / 37
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. . . . . .
A message from the Mathematical Grammar Police
Please do not say “ limx→a
f(x) = DNE.” Does not compute.
I Too many verbs
I Leads to FALSE limit laws like “If limx→a
f(x) DNE and limx→a
g(x) DNE,then lim
x→a(f(x) + g(x)) DNE.”
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 27 / 37
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. . . . . .
A message from the Mathematical Grammar Police
Please do not say “ limx→a
f(x) = DNE.” Does not compute.
I Too many verbsI Leads to FALSE limit laws like “If lim
x→af(x) DNE and lim
x→ag(x) DNE,
then limx→a
(f(x) + g(x)) DNE.”
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 27 / 37
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. . . . . .
Two More Important Limit Theorems
TheoremIf 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 at a),and
limx→a
f(x) = limx→a
h(x) = L,
thenlimx→a
g(x) = L.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 28 / 37
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. . . . . .
Using the Squeeze Theorem
We can use the Squeeze Theorem to replace complicated expressionswith simple ones when taking the limit.
Example
Show that limx→0
x2 sin(πx
)= 0.
SolutionWe have for all x,
−1 ≤ sin(πx
)≤ 1 =⇒ −x2 ≤ x2 sin
(πx
)≤ x2
The left and right sides go to zero as x → 0.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 29 / 37
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. . . . . .
Using the Squeeze Theorem
We can use the Squeeze Theorem to replace complicated expressionswith simple ones when taking the limit.
Example
Show that limx→0
x2 sin(πx
)= 0.
SolutionWe have for all x,
−1 ≤ sin(πx
)≤ 1 =⇒ −x2 ≤ x2 sin
(πx
)≤ x2
The left and right sides go to zero as x → 0.
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. . . . . .
Using the Squeeze Theorem
We can use the Squeeze Theorem to replace complicated expressionswith simple ones when taking the limit.
Example
Show that limx→0
x2 sin(πx
)= 0.
SolutionWe have for all x,
−1 ≤ sin(πx
)≤ 1 =⇒ −x2 ≤ x2 sin
(πx
)≤ x2
The left and right sides go to zero as x → 0.
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. . . . . .
Illustration of the Squeeze Theorem
. .x
.y .h(x) = x2
.f(x) = −x2
.g(x) = x2 sin(πx
)
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. . . . . .
Illustration of the Squeeze Theorem
. .x
.y .h(x) = x2
.f(x) = −x2
.g(x) = x2 sin(πx
)
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 30 / 37
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. . . . . .
Illustration of the Squeeze Theorem
. .x
.y .h(x) = x2
.f(x) = −x2
.g(x) = x2 sin(πx
)
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. . . . . .
Outline
Basic Limits
Limit LawsThe direct substitution property
Limits with AlgebraTwo more limit theorems
Two important trigonometric limits
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. . . . . .
Two important trigonometric limits
TheoremThe following two limits hold:
I limθ→0
sin θθ
= 1
I limθ→0
cos θ − 1θ
= 0
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. . . . . .
Proof of the Sine Limit
Proof.
. .θ
.sin θ
.cos θ
.θ
.tan θ
.−1 .1
Notice
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 left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
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. . . . . .
Proof of the Sine Limit
Proof.
. .θ.sin θ
.cos θ
.θ
.tan θ
.−1 .1
Notice
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 left and right sides tend to 1. So, then, must the middleexpression.
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. . . . . .
Proof of the Sine Limit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
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 left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
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. . . . . .
Proof of the Sine Limit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
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 left and right sides tend to 1. So, then, must the middleexpression.
V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 33 / 37
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. . . . . .
Proof of the Sine Limit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
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 left and right sides tend to 1. So, then, must the middleexpression.
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. . . . . .
Proof of the Sine Limit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
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 left and right sides tend to 1. So, then, must the middleexpression.
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. . . . . .
Proof of the Sine Limit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
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 left and right sides tend to 1. So, then, must the middleexpression.
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. . . . . .
Proof of the Cosine Limit
Proof.
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.
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. . . . . .
Try these
Example
1. limθ→0
tan θθ
2. limθ→0
sin 2θθ
Answer
1. 12. 2
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. . . . . .
Try these
Example
1. limθ→0
tan θθ
2. limθ→0
sin 2θθ
Answer
1. 12. 2
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. . . . . .
Solutions
1. Use the basic trigonometric limit and the definition of tangent.
limθ→0
tan θθ
= limθ→0
sin θθ cos θ
= limθ→0
sin θθ
· limθ→0
1cos θ
= 1 · 11= 1.
2. Change the variable:
limθ→0
sin 2θθ
= lim2θ→0
sin 2θ2θ · 12
= 2 · lim2θ→0
sin 2θ2θ
= 2 · 1 = 2
OR use a trigonometric identity:
limθ→0
sin 2θθ
= limθ→0
2 sin θ cos θθ
= 2 · limθ→0
sin θθ
· limθ→0
cos θ = 2 ·1 ·1 = 2
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. . . . . .
Solutions
1. Use the basic trigonometric limit and the definition of tangent.
limθ→0
tan θθ
= limθ→0
sin θθ cos θ
= limθ→0
sin θθ
· limθ→0
1cos θ
= 1 · 11= 1.
2. Change the variable:
limθ→0
sin 2θθ
= lim2θ→0
sin 2θ2θ · 12
= 2 · lim2θ→0
sin 2θ2θ
= 2 · 1 = 2
OR use a trigonometric identity:
limθ→0
sin 2θθ
= limθ→0
2 sin θ cos θθ
= 2 · limθ→0
sin θθ
· limθ→0
cos θ = 2 ·1 ·1 = 2
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. . . . . .
Solutions
1. Use the basic trigonometric limit and the definition of tangent.
limθ→0
tan θθ
= limθ→0
sin θθ cos θ
= limθ→0
sin θθ
· limθ→0
1cos θ
= 1 · 11= 1.
2. Change the variable:
limθ→0
sin 2θθ
= lim2θ→0
sin 2θ2θ · 12
= 2 · lim2θ→0
sin 2θ2θ
= 2 · 1 = 2
OR use a trigonometric identity:
limθ→0
sin 2θθ
= limθ→0
2 sin θ cos θθ
= 2 · limθ→0
sin θθ
· limθ→0
cos θ = 2 ·1 ·1 = 2
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. . . . . .
Summary
I Limits laws say limits playwell with the rules ofarithmetic
I When limit laws do notwork we can bealgebraically creative
I When algebra does notwork we can trySqueezing.
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