Question

solve this question by l'hospital rule...

Answer 1

We have to find the limit $\lim_{x \to 0} \frac{\log(1-x^2)}{\log \cos x}$.

Since the limit is in $\frac{0}{0}$ form, we can apply L'Hospital's rule.

$\lim_{x \to 0} \frac{\log(1-x^2)}{\log \cos x} = \lim_{x \to 0} \frac{\frac{d}{dx}\log(1-x^2)}{\frac{d}{dx}\log \cos x} \\ \lim_{x \to 0} \frac{\log(1-x^2)}{\log \cos x} = \lim_{x \to 0} \frac{\frac{-2x}{1-x^2}}{-\tan x}$

The limit is again in $\frac{0}{0}$ form, we can apply L'Hospital's rule again,

$\lim_{x \to 0} \frac{\log(1-x^2)}{\log \cos x} = \lim_{x \to 0} \frac{\frac{d}{dx}\frac{-2x}{1-x^2}}{-\frac{d}{dx}\tan x} \\ \lim_{x \to 0} \frac{\log(1-x^2)}{\log \cos x} = \lim_{x \to 0} \frac{\frac{-2}{1-x^2}-\frac{-4x^2}{(1-x^2)^2}}{-\sec^2 x} \\ \lim_{x \to 0} \frac{\log(1-x^2)}{\log \cos x} = \frac{\frac{-2}{1-0}-\frac{-0}{(1-0)^2}}{-\sec^2 0} \\ \lim_{x \to 0} \frac{\log(1-x^2)}{\log \cos x} = 2$