How do you use the Squeeze Theorem to show that $limsinx/x$ as x approaches infinity?

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Answer 1 · 2015-09-27T04:04:38

Refer to explanation

Hence

$-1<=sinx<=1=>-1/x<=sinx/x<=1/x$

Hence $lim_(x->oo) -1/x=0$ and $lim_(x->oo) 1/x=0$

so $lim_(x->oo) sin/x=0$

How do you use the Squeeze Theorem to show that limsinx/xlimsinx/x as x approaches infinity?