# lim (cos x )^(1/x ) when x = 0

#### xl5899

here's my working , is it correct ?

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#### Prove It

MHF Helper
There's a small typo, and you only have to use L'Hospital's Rule once.

\displaystyle \begin{align*} \lim_{x \to 0} \frac{\ln{ \left[ \cos{(x)} \right] } }{ x } &= \lim_{x \to 0} \frac{-\frac{\sin{(x)}}{\cos{(x)}}}{1} \\ &= \lim_{x \to 0} \left[ -\tan{(x)} \right] \\ &= -\tan{(0)} \\ &= 0 \end{align*}

#### Archie

You don't have to use l'Hôpital at all. Instead we note that $$\displaystyle \lim_{x \to 0}\left(1+x\right)^{1 \over x}=\matjrm e$$. Use the double angle formula $$\displaystyle 2\cos^2 x= 1+\cos 2x$$ or the first two terms of the MacLaurin series for $$\displaystyle \cos x$$ to get the correct form inside the brackets.