Question : Evaluate
$\displaystyle
\lim_{x \to 0} \ (Cos x)^{\frac{1}{x^2}}
$
as $\displaystyle x\to0$ we know that $\displaystyle \frac{\ln(1-x)}x\to-1$ and $\displaystyle \frac{1-\cos x}{x^2}\to\frac12.$
now following up on mr fantastic work, we have that for $\displaystyle x\ne0$ is $\displaystyle \frac{\ln (\cos x)}{x^{2}}=\frac{\ln \big(1-(1-\cos x)\big)}{1-\cos x}\cdot \frac{1-\cos x}{x^{2}},$ and as $\displaystyle x\to0$ the whole limit is $\displaystyle \frac1{\sqrt e}.$
$\displaystyle \lim_{x \to 0} \frac{ln \ cosx}{x^2}$ = $\displaystyle \lim_{x \to 0} \frac{f(x)}{g(x)}$
Using L'Hospitals Rule weget
$\displaystyle \lim_{x \to 0} \frac{f(x)}{g(x)}$ = $\displaystyle \lim_{x \to 0} \frac{f'(x)}{g'(x)}$
$\displaystyle \lim_{x \to 0} \frac{f'(x)}{g'(x)}$ = $\displaystyle \lim_{x \to 0} \frac{- tanx}{2x}$
Dividing nu & de by x weget
= $\displaystyle \lim_{x \to 0} \frac{- \frac{tanx}{x}}{2}$
= $\displaystyle - \frac{ 1}{2}$
Is this correct or have i done something wrong.....And after this what should i do??