estimate the limit of $(\cos(4x))^\frac{1}{x^2}$ I was looking for a way to estimate the limit of $\left(\cos\left(4x\right)\right)^\frac{1}{x^2}$ as x goes to zero using lohospital laws, which I didn't consider to be possible given there is no 0/0 here...how can one do this using lhospital? 
 A: hint: consider $$\lim_{x \to 0}(\cos(4x))^{\frac{1}{x^2}}=\exp\left(\lim_{x \to 0}\frac{\ln(\cos(4x))}{x^2}\right)$$
A: Firstly,
$$(\cos(4x))^\frac{1}{x^2}=e^{\frac{\ln(\cos(4x))}{x^2}}.$$
Then,
$$\lim_{x\to 0}\frac{\ln(\cos(4x))}{x^2}\underset{Hop.}{=}\lim_{x\to 0}\frac{-4\sin(4x)}{2\cos(4x)x}=\lim_{x\to 0}-2\cdot \underbrace{\frac{\sin(4x)}{4x}}_{\to 1}\cdot \frac{4}{\cos(4x)}=-8.$$
Moreover, $e^x$ is continuous on $x=-8$, therefore
$$\lim_{x\to 0}e^{\frac{\ln(\cos(4x))}{x^2}}=e^{\lim_{x\to 0}\frac{\ln(\cos(4x))}{x^2}}=e^{-8}.$$
A: It is well-known that, if $ \lim f(x) = \exp \left( \lim \ln f(x) \right) $, by continuity. Hence, $$\lim_{x \to 0}(\cos(4x))^{\frac{1}{x^2}}=\exp\left(\lim_{x \to 0}\frac{\ln(\cos(4x))}{x^2}\right)$$Now, using l'Hopital's rule gives the inner limit to be $$ \begin {align*} \lim_{x \to 0} \frac {\frac{-4 \sin (4x)}{x\cos(4x)}}{2x} &= -2 \cdot \lim_{x \to 0} \frac {\sin (4x)}{x \cos (4x)} \\&= -2 \cdot \lim_{x \to 0} \frac {\tan(4x)}{x} \\&= -2 \cdot \lim_{x \to 0} \frac {4\sec^2(4x)}{1} \\&= -8 \cdot \sec^2 (0) = -8. \end {align*} $$Hence, the answer is $e^{-8}$. $\Box$
