Question on limits $ \lim\limits_{x\to 0} (\frac{\cos x}{\cos2x})^{\frac{1}{x^2}}$ $$ \lim_{x\to 0} \left(\frac{\cos x}{\cos2x}\right)^{\frac{1}{x^2}}$$
so this is my first time using this, hope i typed it out correctly.
Could anyone give me a hint on this? i do know that $\cos2x= 2(\cos x)^2-1$ but does that help in solving the question?
Probable ans?:
$ \lim_{x\to 0} \left(\frac{\cos x}{\cos2x}\right)^{\frac{1}{x^2}}$
=$ exp\lim_{x\to 0}\left(\frac{\ln\left(\frac{cosx}{cos2x}\right)}{x^2}\right)$
=$ exp\lim_{x\to 0}\left(\frac{\left(\frac{2sin2x}{cos2x}-\frac{sinx}{cosx}\right)}{2x}\right)$
=$ exp\lim_{x\to 0}\left(\frac{\left(2tan2x-tanx\right)}{2x}\right)$
=$ exp\lim_{x\to 0}\left(\frac{\left(4(secx)^2-(secx)^2\right)}{2}\right)$
=$exp(\frac32)$
i applied L'Hôpital's rule on lines 3 & 4. Would this be the correct answer?
 A: HINT:
For $m^2-n^2\ne0,$
$$\left(\frac{\cos2mx}{\cos2nx}\right)^{\dfrac1{x^2}}=\left[\left(1+\frac{\cos2mx-\cos2nx}{\cos2nx}\right)^{\dfrac{\cos2mx}{\cos2mx-\cos2nx}}\right]^{\dfrac{\cos2mx-\cos2nx}{x^2}}$$
For the inner limit, $\lim_{h\to0}\left(1+h\right)^{\dfrac1h}=\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$
Now using Prosthaphaeresis Formula, $\cos2mx-\cos2nx=2\sin(m+n)x\cdot\sin(n-m)x$
and finally use $\lim_{u\to0}\dfrac{\sin(ru)}u=r\cdot\lim_{u\to0}\dfrac{\sin(ru)}{ru}=r$
to find the limit to be $\displaystyle e^{2(n^2-m^2)}$
Here $2m=1,2n=2$
Hope you can take it home from here
A: Let $$A= \left(\frac{\cos(x)}{cos(2x)}\right)^{\frac{1}{x^2}}$$ So, as already suggested $$x^2 \log(A)=\log\Big(\frac{\cos(x)}{\cos(2x)}\Big)=\log\Big(\cos(x)\Big)-\log\Big(\cos(2x)\Big)$$ Now consider the Taylor expansion of $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^5\right)$$ So $$\log\Big(\cos(y)\Big)\approx\log\Big(1-(\frac{y^2}{2}-\frac{y^4}{24})\Big)\approx -\frac{y^2}{2}-\frac{y^4}{12}$$ Replacing $y$ by $x$ and then $y$ by $2x$ for the second logarithm then gives $$x^2 \log(A)\approx\frac{3 x^2}{2}+\frac{5 x^4}{4}$$
I am sure that you can take from here.
A: HINT:
$\cos x\approx 1-\frac{x^2}2$, hence $\cos 2x\approx 1-2{x^2}$. Now
$$
\lim_{x\to0}(\cos x)^{1/x^2}=e^{-1/2},\quad
\lim_{x\to0}(\cos 2x)^{1/x^2}=e^{-2},
$$
and the limit is $e^{3/2}$.
