Find $\lim_{x\to 0^+}\frac{1}{x^2}\int_{0}^{2x}\ln(\cos t) dt$

Question

$$ \lim_{x\to0^+}\frac{1}{x^2}\int_{0}^{2x}\ln(\cos t) \,dt$$

$F(x)=\int_{0}^{2x}\ln(\cos t) \,dt=\ln(\cos 2x)$

$ \lim_{x\to0^+}\frac{1}{x^2}\int_{0}^{2x}\ln(\cos t)\, dt= \lim_{x\to0^+}\frac{\ln(\cos 2x)}{x^2}=-2$

Is valid my answer?

Answer 1

$$ \ln(\cos t)=\ln\Big(1-\frac{t^2}{2}+{\mathcal O}(t^4)\Big)=-\frac{t^2}{2}+{\mathcal O}(t^4), $$ and hence $$ \int_0^{2x}\ln(\cos t)\,dt=-\frac{8x^3}{6}+{\mathcal O}(x^5) $$ and $$ \frac{\int_0^{2x}\ln(\cos t)\,dt}{x^2}=-\frac{8x}{6}+{\mathcal O}(x^3)\to 0. $$

Answer 2

$lim_{x \rightarrow 0+} \frac{\int_{0}^{2x}ln(cos t)dt}{x^2}$ $[\frac{0}{0}$ form]

=$lim_{x \rightarrow 0+} \frac{ln(cos 2x)}{x}$ [Taking the derivative][$\frac{0}{0}$ form]

=$lim_{x \rightarrow 0+} \frac{- sin 2x}{cos 2x}\cdot 2$

=$0$