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

Question

4828 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-22T22:52:59

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

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

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

So,

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

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