# How do you find the limit of (cos (2x))^(3/(x^2)) as x approaches 0?

Mar 4, 2015

The limit should be $\frac{1}{e} ^ 6$.
${\lim}_{x \to 0} {\cos}^{\frac{3}{x} ^ 2} \left(2 x\right) =$

But:
cos^(3/x^2)(2x)=e^[3/x^2ln[cos(2x)] (have a look at the properties of logarithms)

and:
${\lim}_{x \to 0} {e}^{\frac{3}{x} ^ 2 \ln \left[\cos \left(2 x\right)\right]} = {e}^{-} 6$

The exponent $\frac{3}{x} ^ 2 \ln \left[\cos \left(2 x\right)\right]$ tends to $- 6$:

hope it is clear