Question

Question #0448b

Answer 1

`1`

Explanation:

`lim_(xrarr0)sin^2(x)^2/x^4 = lim_(xrarr0) (sinx^2/x^2 * sinx^2/x^2)`

`= lim_(xrarr0) (sinx^2/x^2) * lim_(xrarr0)(sinx^2/x^2)`

`= 1*1 = 1`

Note

`lim_(xrarr0)(sinx^2/x^2)=1` by using the fundamental trigonometric limit

`lim_(thetararr0) sintheta/theta = 1`

with `theta = x^2`.