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Asked Feb 7, 2020
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I just need reassurance.  Because x is approaching 0 that means it cannot equal 0 but be anything greater than or equal to .  So why couldn't I say greater than 0 or less than 0?

5. Find the following limit and explain which limit laws you used:
lim x2008 sin
+ cos x
r2009
Dind the following limit andl ovnlein which limit lowa vou usod:
lim £lo) whore £(.
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5. Find the following limit and explain which limit laws you used: lim x2008 sin + cos x r2009 Dind the following limit andl ovnlein which limit lowa vou usod: lim £lo) whore £(.

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5. First, we note that, for x + 0
-1 < sin(
+ cos x) < 1.
2009
Hence, for x 0, we have
2008
'<²sin(-
x2008
,2008
(0.1)
+ cos x) < “
We know that
2008
lim -x
and
lim x2008 = 0.
Therefore, from (0.1) and the Squeeze Theorem, we conclude that
(
-)=
2008 sin
lim x
0.
+ cos x
-2009
help_outline

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5. First, we note that, for x + 0 -1 < sin( + cos x) < 1. 2009 Hence, for x 0, we have 2008 '<²sin(- x2008 ,2008 (0.1) + cos x) < “ We know that 2008 lim -x and lim x2008 = 0. Therefore, from (0.1) and the Squeeze Theorem, we conclude that ( -)= 2008 sin lim x 0. + cos x -2009

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Expert Answer

Step 1

Squeeze theorem states that

Calculus homework question answer, step 1, image 1

Step 2

Consider,

Calculus homework question answer, step 2, image 1

Step 3

Calculus homework question answer, step 3, image 1

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