Need help to solve this limit (Calculus) without L'Hospital's Rule

$$\lim_{x \to 0} \dfrac{\log(\cos x)}{\tan (x^2)}$$

That's the limit I need to solve, please notice that I can't use L'Hospital's Rule, now I'll show you what I've done so far:

$$\lim_{x \to 0} \dfrac{\log(\cos x)}{\tan (x^2)} = \lim_{x \to 0} \dfrac{\log (\cos x)}{\dfrac{\sin (x^2)}{\cos (x^2)}} = \dfrac{\cos (x^2) \log(\cos x)}{\sin (x^2)}$$

I was thinking about transforming the $\sin x$ into a $\cos x$ using this identity $\sin(x + \pi/2) = \cos x$, but it hasn't worked so far.

• By $\tan x^2$ do you mean $\tan\left(x^2\right)$ or $\tan(x)^2$? The latter is often written as $\tan^2(x)$ – robjohn Jan 21 '18 at 22:41
• Thanks for pointing that out, I've already edited the question so that it's clearer. – M. Navarro Jan 21 '18 at 22:52

Note that

$$\frac{\log \cos x}{\tan x^2}=\frac12\frac{\log \cos^2 x}{\tan x^2}=\frac12\frac{\log (1-\sin^2 x)}{\tan x^2}=\\=\frac12\frac{\log (1-\sin^2 x)}{\sin x^2}\frac{\sin x^2}{x^2}\frac{x^2}{\tan x^2}\to\frac12\cdot-1\cdot1\cdot1=-\frac12$$

since the following standard limits hold

$$\frac{\sin x}{x}\to 1 \quad\frac{\tan x}{x}\to 1\quad \frac{\log (1+x)}{x}\to 1\implies -\frac{\log (1-x)}{-x}\to -1$$

Hint:

First observe $\tan x^2\sim_0 x^2$.

Then compose Taylor's expansion at order $2$ of $\cos x$ with Taylor's expansion of $\log(1+u)$ at order $1$ to obtain an equivalent of the numerator.