Question

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

Answer 1

Refer to explanation

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`