Question

Find the limit as x approaches infinity of `y=sinx/x`?

Answer 1

`lim_(x->oo) sinx/x = 0`

Explanation:

You're going to want to use the squeeze theorem for this. Recall that `sinx` is only defined on `-1 ≤ sinx ≤ 1`. Therefore

`-1/x ≤ sinx/x ≤ 1/x`

And since `lim_(x-> oo) -1/x = lim_(x-> oo) 1/x = 0`, then `lim_(x->oo) sinx/x = 0`.

Hopefully this helps!