Question

How do you use the Squeeze Theorem to find `lim ((sin n )^2) / 2^n` as x approaches zero?

Answer 1

Assuming that the question should be `lim ((sin x )^2) / 2^x` as `xrarr0`, you don't need the squeeze theorem for that limit.

Explanation:

`lim_(xrarr0)(sinx)^2 = 0^2 =0`

`lim_(xrarr0)2^x = 2^0 = 1`

So,

`lim ((sin x )^2) / 2^x = 0/1 =0`