# 5. Find the following limit and explain which limit laws you used:lim x2008 sin+ cos xr2009Dind the following limit andl ovnlein which limit lowa vou usod:lim £lo) whore £(. 5. First, we note that, for x + 0-1 < sin(+ cos x) < 1.2009Hence, for x 0, we have2008'

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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? help_outlineImage Transcriptionclose5. 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 £(. fullscreen help_outlineImage Transcriptionclose5. 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 fullscreen
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Step 1

Squeeze theorem states that Step 2

Consider, Step 3 ...

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