Suppose we wish to evaluate limx→0x10sin(2x)+8.Since we know that −1≤sin(θ)≤1 for θ∈R, we must haveCorrect ≤sin(2x)≤ Correct for x≠0. Part 2 of 4 From this, we must have   ≤x10sin(2x)≤   for x≠0.  ≤x10sin(2x)+8≤   for x≠0.

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Suppose we wish to evaluate limx→0x10sin(2x)+8.

Since we know that −1≤sin(θ)≤1 for θ∈R, we must have

Correct ≤sin(2x)≤ Correct for x≠0.

Part 2 of 4

From this, we must have

  ≤x10sin(2x)≤   for x≠0.

  ≤x10sin(2x)+8≤   for x≠0.
 
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