Please calculate $\lim_{x\rightarrow 0}\frac{\ln(\cos x)}{x^2}$ $$\lim_{x\rightarrow0}\frac{\ln(\cos x)}{x^2}$$
I've tried this 
$$\lim_{x\rightarrow0}\left(\frac{\ln(\cos x)+1}{\cos x}\cdot\frac{\cos x}{x^{2}}-\frac{1}{x^{2}}\right)$$
but $\dfrac{1}{x^2}$ is goint to infinity. Please answer, how to solve it using easy methods. Thanks in advance!
 A: Hint#1:
You can use these series (you can find them here)
$$\cos(x)\approx1-\frac{x^2}{2}+o(x^2)$$
$$\ln(1-x)\approx -x+o(x)$$
Here I used O notation
After the consistent application of these series, you will get the correct answer $-\frac{1}{ 2}$
Edit
Hint#2
$$\ln(\cos(x))\approx\ln\left(1-\frac{x^2}{2}+o(x^2)\right)\approx-\frac{x^2}{2}+o\left(-\frac{x^2}{2}+o(x^2)\right)\approx-\frac{x^2}{2}+o(x^2)$$
All properties of $o(f)$ you can find here (in section "Little-o notation")
A: First write $\ln(\cos x)$ as $\frac{1}{2}\ln(1-\sin^2 x)$, then see the when $x\to 0$,  $\sin^2 x$ is very small. We know that when the function $\alpha(x)$ is very small, then $\ln(1+\alpha(x))\sim\alpha(x)$ so $$\ln\left(\cos x\right)=\frac{1}{2}\ln(1+(-\sin^2 x))\sim \frac{-1}{2}\sin^2(x)$$
Now take your limit with this fact again. 
A: (using L'Hopital rule)
$$\lim_{x\rightarrow0}\frac{\ln(\cos x)}{x^2}=\lim_{x\to 0}\frac{-\tan x}{2x}=\lim_{x\to 0}\frac{-1}{2\cos^2x}=-\frac{1}{2}$$
A: Solution with no De-L'Hopital, Taylor Expansions or assymptotic equalities but assuming the limit exists:
First prove that
$$\lim_{x\rightarrow0}\frac{\ln(\cos x)}{x}=0$$
This follows from the fact that
$$\lim_{x\rightarrow0}\frac{\ln(\cos x)}{x}=\lim_{x\rightarrow0}\frac{\ln(\cos x)-\ln(\cos 0)}{x}=(\ln(\cos x))^{\prime}(0)$$
Now define 
$$f(x)=\begin{cases}\frac{\ln(\cos x)}{x}& x\neq 0\\
0& x=0\end{cases}$$
$f$ is continuous everywhere and differentiable at least in $\mathbb{R}^*$
Observe that
$$\lim_{x\rightarrow0}\frac{\ln(\cos x)}{x^2}=\lim_{x\to 0}\frac{f(x)-f(0)}{x}$$
and so we need to evaluate $f^{\prime}(0)$ (after proving its existence).
We state:
$$\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}f^{\prime}(x)$$
Indeed, by the Mean Value Theorem for $x>0$, $\exists \xi_x\in (0,x)$ so that $$f^{\prime}(\xi_x)=\frac{f(x)-f(0)}{x}$$
Letting $x\to 0^+$, $\xi\to 0^+$ and so  $$\lim_{x\to 0^+}f^{\prime}(x)=\lim_{x\to 0^+}\frac{f(x)-f(0)}{x}$$
Similarly for $x<0$ (the above is done assuming $\lim_{x\to 0}f^{\prime}(x)$ exists). 
It remains to show $\lim_{x\to 0}f^{\prime}(x)$ exists and to evaluate it. Showing existence will not be trivial as the limit $\lim_{x\to 0}\frac{\cos x}{x^2}$ re-appears, albeit with a minus sign. Assuming the existence of $\lim_{x\to 0}\frac{\cos x}{x^2}$ however, one can easily evaluate it as
$$\lim_{x\to 0}\frac{\cos x}{x^2}=\lim_{x\to 0}f^{\prime}(x)$$
(the last limit will have a term $-\lim_{x\to 0}\frac{\cos x}{x^2}$ and another easy limit. Pair up $\lim_{x\to 0}\frac{\cos x}{x^2}$ and you will be done
Of course if the existence is not assumed, one would have to use other trickery.
