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

Mar 2014
909
2
malaysia
here's my working , is it correct ? DSC_0091[1].JPG
 
Last edited:

Prove It

MHF Helper
Aug 2008
12,897
5,001
There's a small typo, and you only have to use L'Hospital's Rule once.

$\displaystyle \begin{align*} \lim_{x \to 0} \frac{\ln{ \left[ \cos{(x)} \right] } }{ x } &= \lim_{x \to 0} \frac{-\frac{\sin{(x)}}{\cos{(x)}}}{1} \\ &= \lim_{x \to 0} \left[ -\tan{(x)} \right] \\ &= -\tan{(0)} \\ &= 0 \end{align*}$
 
Dec 2013
2,002
757
Colombia
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.