# How do you find the limit of (cos x)^(1/x^2) as x approaches 0?

Jun 3, 2016

$\frac{1}{\sqrt{e}}$

#### Explanation:

$\left[1\right] \text{ } {\lim}_{x \to 0} {\left(\cos x\right)}^{\frac{1}{x} ^ 2}$

This is an indeterminate form of the type ${1}^{\infty}$. You need to first convert it to the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ so you can use L'Hopital's Rule. We can do this by using $e$ and $\ln$.

$\left[2\right] \text{ } = {\lim}_{x \to 0} {e}^{\ln} \left[{\left(\cos x\right)}^{\frac{1}{x} ^ 2}\right]$

$\left[3\right] \text{ } = {\lim}_{x \to 0} {e}^{\left(\frac{1}{x} ^ 2\right) \ln \left(\cos x\right)} = {\lim}_{x \to 0} {e}^{\ln \frac{\cos x}{x} ^ 2}$

We can take out $e$.

$\left[4\right] \text{ } = {e}^{{\lim}_{x \to 0} \ln \frac{\cos x}{x} ^ 2}$

This is now an indeterminate form of the type $\frac{0}{0}$. We can use L'Hopital's Rule now. Get the derivatives of both the numerator and denominator.

[5]" "=e^(lim_(x->0)(-sinx/cosx)/(2x))=e^(lim_(x->0)(-sinx/(2xcosx))

This is still indeterminate so you must apply L'Hopital's Rule again.

[6]" "=e^(lim_(x->0)(-cosx/(2(-xsinx+cosx)))

You can now get the limit by substitution.

$\left[6\right] \text{ } = {e}^{- \cos \frac{0}{2 \left(- 0 \sin 0 + \cos 0\right)}}$

$\left[7\right] \text{ } = {e}^{- \frac{1}{2}}$

$\left[8\right] \text{ } = \textcolor{b l u e}{\frac{1}{\sqrt{e}}}$