Find $\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$$
What I have done is to take the $\ln$ 
$$e^{\lim_{x \to 0}\ln\left(\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}\right)}$$
$$y={\lim_{x \to 0}{\frac{1}{x^2}}\cdot \ln\frac{\cos x}{\cos(2x)}}$$
Applying l'hopital to get 
$$y={\lim_{x \to 0}{\frac{-2}{x^3}}\cdot \frac{\cos(2x)}{\cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x)^2)}}={\lim_{x \to 0}\frac{-2}{x^3\cdot \cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x))}}$$
Should I apply l'hopital again?

If you start from $g(x) = \frac{\cos x}{\cos(2x)}^{\frac{1}{x^2}}$ and first take logs you have $\ln g(x) = {\frac{1}{x^2}} (\ln \frac{\cos x}{\cos(2x)})$
Expanding $\frac{\cos(x)}{\cos(2x)} \approx 1 + 3/2 x^2$ and $\ln (1+x) \approx  x$ you have $\ln g(x)  \approx \frac{3/2 x^2}{x^2} = 3/2$, so that the limit is $e^{3/2}$.
