Limit of real logarithm During my homework, I've got to the following exercise:
Calculate the following limit (if exists):
$$\lim \limits_{x \to 0}\left(\frac{\ln\big(\cos(x)\big)}{x^2}\right).$$ 
I looked in the Calculus book to find an identity that will help me resolve the expression, and what I found was:
$$\frac{1}{n}\ln(n) = \ln(\sqrt[n]n)$$
I couldn't manage to prove that Identity and thus couldn't solve the question.
So what I need is:


*

*Proof to the identity,

*Help in solving the question.


Context:
Didn't learn either L'Hopital, McLauren or Taylor Series yet.
This question came just after the "Continuity" chapter.
 A: If you can use:
$$\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1$$
(which can be easily proven by considering the function $u\mapsto \ln(1+u)$, and showing that it is differentiable at $0$ with derivative equal to $1$)
and
$$\lim_{u\to 0}\frac{\sin u}{u} = 1$$
which also can be shown the same way $\sin^\prime 0 = \cos 0 = 1$), 
then you have that, for $x\in (-\frac{\pi}{2}, \frac{\pi}{2})\setminus\{0\}$:
$$
\frac{\ln\cos x}{x^2} = \frac{\ln \sqrt{1-\sin^2 x}}{x^2} = \frac{1}{2}\frac{\ln(1-\sin^2 x)}{x^2} = -\frac{1}{2}\frac{\ln(1-\sin^2 x)}{-\sin^2 x}\cdot\frac{\sin^2 x}{x^2}. 
$$
Using the above, since $\sin^2 x\xrightarrow[x\to0]{} 0$ , you get, as all limits exist:
$$
\frac{\ln\cos x}{x^2} \xrightarrow[x\to0]{} -\frac{1}{2}\left(\lim_{u\to 0}\frac{\ln (1+u)}{u}\right)\cdot \left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2 = -\frac{1}{2}
$$
A: Use Taylor's formula at order  $2$ and equivalents:


*

*$\cos x=1-\dfrac{x^2}2+o(x^2)$

*$\ln(1-u)\sim_0 -u$, hence $\ln(\cos x)\sim_0 -\dfrac{x^2}2+o(x^2)\sim_0 -\dfrac{x^2}2$.


From this, we deduce
$$\frac{\ln\cos x}{x^2}\sim_0 -\frac{x^2}{2x^2}=-\frac12.$$
Without Taylor's formula:
You can prove directly that $\;\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{\sin^2 x }{x^2(1+ \cos x)}=\frac12$, so that
$$\frac{1-\cos x}{x^2}\sim_0\frac12,\quad\text{whence}\quad \cos x\sim_0 1 - \dfrac{x^2}2.$$
A: There are many ways to solve this question:


*

*have you heard about Taylor series? By applying McLaurin rules you get
$$\lim \limits_{x \to 0}\frac{\ln(\cos(x))}{x^2} = \lim \limits_{x \to 0}\frac{\ln \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^4)\right)}{x^2}\\=\lim \limits_{x \to 0}\frac{-\frac{x^2}{2!} + O(x^2)}{x^2} = \lim \limits_{x \to 0}\frac{x^2 \cdot \left( -\frac{1}{2} + \frac{O(x^2)}{x^2}\right)}{x^2} = \lim \limits_{x \to 0}-\frac{1}{2} + \frac{O(x^2)}{x^2} = -\frac{1}{2}$$

*another approach is using L'Hospital's rule:
differentiate both numerator and denominator (given you checkes the hypothesis) and you get
$$\lim \limits_{x \to 0}\frac{\ln(\cos(x))}{x^2} = \lim \limits_{x \to 0}\frac{\frac{1}{\cos x} \cdot -\sin x}{2x}$$ and since $ \lim \limits_{x \to 0}\frac{\sin x}{x} = 1$ you can simplify to get
$$$\lim \limits_{x \to 0}\frac{\ln(\cos(x))}{x^2} = \lim \limits_{x \to 0}\frac{\frac{1}{\cos x} \cdot -\sin x}{2x} = \lim \limits_{x \to 0}-\frac{1}{2 \cos x} = -\frac{1}{2}$$

*classic approach is to check the limit by its definition: let $f(x) = \frac{\ln(\cos(x))}{x^2}$, what you want to check is that for every number $\epsilon$ there is some number $\delta$ such that
$$|f(x) - \left( -\frac{1}{2} \right) | < \epsilon$$
whenever $| x - 0| < \delta$

A: Use the well-known Taylor series 
$$\cos(x)= 1-\frac{1}{2}x^2+\frac{1}{24}x^4+O(x^6)$$
$$\ln(1+x)= x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5+O(x^6)$$
to get 
$$\ln(\cos(x))=-\frac{1}{2}x^2-\frac{1}{12}x^4+O(x^6)$$
Can you continue?
