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?
hint: consider $$\lim_{x \to 0}(\cos(4x))^{\frac{1}{x^2}}=\exp\left(\lim_{x \to 0}\frac{\ln(\cos(4x))}{x^2}\right)$$
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 $$ \begin {align*} \lim_{x \to 0} \frac {\frac{-4 \sin (4x)}{x\cos(4x)}}{2x} &amp;= -2 \cdot \lim_{x \to 0} \frac {\sin (4x)}{x \cos (4x)} \\&amp;= -2 \cdot \lim_{x \to 0} \frac {\tan(4x)}{x} \\&amp;= -2 \cdot \lim_{x \to 0} \frac {4\sec^2(4x)}{1} \\&amp;= -8 \cdot \sec^2 (0) = -8. \end {align*} $$Hence, the answer is $e^{-8}$. $\Box$