Question

How do you find the limit of `sin (x^2)/sin^2(2x)` as x approaches 0?

Answer 1

`1/4`

Explanation:

We will be using

`lim_(x->0)sinx/x=1` so

`sin (x^2)/sin^2(2x)=((4x^2)/(4x^2))sin (x^2)/sin^2(2x)=1/4((sin(x^2)/x^2))/((sin(2x)/(2x))^2)` then

`lim_(x->0)sin (x^2)/sin^2(2x)=lim_(x->0)1/4((sin(x^2)/x^2))/((sin(2x)/(2x))^2)=1/4 1/1 = 1/4`