Evaluating $\lim_{x \to 0} \frac{ x^2-\ln(\cos{x})}{x} $

Question

$$\lim_{x \to 0} \frac{ x^2-\ln(\cos{x})}{x}$$

Our domain is $]-π/2,π/2[$ the original question is :

$F(x) = x^2-\ln(\cos x)$

$F'(x) = 2x+\tan(x)$

And we want to know $\lim_{x\to 0} \frac{f(x)}{x}$.

Answer 1

For the highlighted question use L'Hopital's Rule. Since $\lim \frac {2x+tan\, x} 1=0$ the answer is $0$.

Answer 2

Without L'Hopital: let $f(x):= x^2- \ln ( \cos x).$ Then

$$\lim_{x \to 0} \frac{ x^2-\ln(\cos{x})}{x}= \lim_{x \to 0} \frac{ f(x)-f(0)}{x-0}=f'(0)=0.$$