Find the limit of $\lim_{x\to 0}\left(\frac{\cos x}{\cos 2x}\right)^\frac{1}{x^2}$ Find limit of $$\lim_{x\to 0}\left(\frac{\cos x}{\cos2x}\right)^{x^{-2}}.$$
After some algebra I get $$\lim_{x\to 0}\exp\left(\left(\frac{\cos x-2(\cos x)^2+1}{\cos2x}\right)\cdot\frac{1}{x^2}\right).$$ Maybe I am missing something but I can't continue from this.
 A: Hint:
Let $y = \left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}$.
We need to find $\lim_{x\to 0} y$, however, this might get super ugly.
So instead, try to compute $\lim_{x\to 0} \ln(y) = \lim_{x\to 0} \frac{1}{x^2} \ln\left(\frac{\cos(x)}{\cos(2x)}\right) = \lim_{x\to 0} \frac{\ln\left(\frac{\cos(x)}{\cos(2x)}\right) }{x^2}$.
You will have to use L-Hospital twice, and it gets messy algebraically, but very much doable. Then say your limit is $\alpha$, now since $\ln$ is continuous, we know that $\lim_{x \to 0} y = e^{\alpha}$
A: Rearrange the latter form of the limit that you are getting so that the limit becomes
\begin{equation}
\lim_{x\to 0}\exp\left(\left(\frac{\cos x-2(\cos x)^2+1}{\cos2x}\right)\cdot\frac{1}{x^2}\right)=\exp\left(\lim_{x\to 0}\dfrac{1-\cos x}{x^2}\times\lim_{x\to0}\dfrac{1+2\cos x}{\cos2x}\right)=\boxed{e^{\frac{3}{2}}}
\end{equation}
A: First of all $$\lim _{x\to 0}\left(\left(\frac{cosx\:}{cos2x\:}\right)^{\frac{1}{x^2}}\right)=\lim _{x\to 0}\left(e^{\frac{ln\left(\frac{cosx}{cos2x}\right)}{x^2}}\right)=e^{\lim _{x\to 0}\left(ln\left(\frac{cosx}{cos2x}\right)\right)}$$
Now, we will calculate:
$$\lim _{x\to 0}\left(\frac{ln\left(\frac{cosx}{cos2x}\right)}{x^2}\right)=\lim \:_{x\to \:0}\left(\frac{ln\left(cosx\right)-ln\left(cos2x\right)}{x^2}\right)=\lim \:\:_{x\to \:\:0}\left(\frac{-tanx+2tan2x}{2x}\right)=\lim \:\:_{x\to \:\:0}\left(\frac{4sec^2x-sec^{2x}}{2}\right)=\frac{3}{2}$$
So the final answer is $e^{\frac{3}{2}}$
A: I thought it might be instructive to present an approach that relies on straightforward arithmetic and two standard limits and circumvents use of logarithms, L'Hospital's Rule, and Taylor's Theorem.  To that end we now proceed.

We begin by writing the term of interest as
$$\begin{align}
\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}&=\left(\frac{2\cos^2(x)-1}{\cos(x)}\right)^{-1/x^2}\\\\
&=\left(1-\underbrace{\left(\frac{1-\cos(x)}{x^2}\right)\left(1+\frac{1+\cos(x)}{\cos(x)}\right)}_{\text{denote this as}\,f(x)}x^2\right)^{-1/x^2}\\\\
&=\left(\left(1-f(x)x^2\right)^{-1/f(x)x^2}\right)^{f(x)}
\end{align}$$
Noting that $\lim_{x\to 0}f(x)=3/2$ so that $\lim_{x\to 0}f(x)x^2=0$, and using $\lim_{t\to 0}\left(1-t\right)^{-1/t}=e$, we conclude that
$$\lim_{x\to 0}\left(\frac{\cos(x)}{\cos(2x)}\right)=e^{3/2}$$
NOTE:  We tacitly used the "standard limit" $\displaystyle \lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac12$
A: For each $x\ne0$,$$\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}=\exp\left(\frac{\log\left(\frac{\cos(x)}{\cos(2x)}\right)}{x^2}\right)$$and\begin{align}\lim_{x\to0}\frac{\log\left(\frac{\cos(x)}{\cos(2x)}\right)}{x^2}&=\lim_{x\to0}\frac{-\tan(x)+2\tan(2x)}{2x}\\&=\frac12\left(-\lim_{x\to0}\frac{\tan x}x+2\lim_{x\to0}\frac{\tan(2x)}x\right)\\&=\frac32.\end{align}Therefore$$\lim_{x\to0}\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}=e^{3/2}.$$
A: Hint
Let
$$1+y=1+\dfrac{\cos x-\cos2x}{\cos2x}=1+\dfrac{2\sin\dfrac x2\sin\dfrac{3x}2}{\cos2x}$$
$$(1+y)^{1/x^2}=((1+y)^{1/y})^{y/x^2}$$
Now use $\lim_{y\to0}(1+y)^{1/y}=e$
and $\lim_{h\to0}\dfrac{\sin h}h=1$
