$$\lim_{x \to 0} \dfrac{\log(\cos x)}{\tan (x^2)}$$
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.
Thanks in advance.
Answer
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
$$\frac{\sin x}{x}\to 1 \quad\frac{\tan x}{x}\to 1\quad \frac{\log (1+x)}{x}\to 1\implies -\frac{\log (1-x)}{-x}\to -1$$
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