# Question #004d9

Feb 26, 2017

${\lim}_{x \rightarrow 2} \frac{\cos \left(\frac{\pi}{x}\right)}{x - 2} = \frac{\pi}{4}$

#### Explanation:

${\lim}_{x \rightarrow 2} \frac{\cos \left(\frac{\pi}{x}\right)}{x - 2}$

Attempting to evaluate the limit gives $\cos \frac{\frac{\pi}{2}}{2 - 2} = \frac{0}{0}$, which is an indeterminate form. Thus, we can use l'Hopital's rule, which says to take the derivative of the numerator and denominator separately:

$= {\lim}_{x \rightarrow 2} \frac{\frac{d}{\mathrm{dx}} \left(\cos \left(\frac{\pi}{x}\right)\right)}{\frac{d}{\mathrm{dx}} \left(x - 2\right)}$

$= {\lim}_{x \rightarrow 2} \frac{- \sin \left(\frac{\pi}{x}\right) \frac{d}{\mathrm{dx}} \left(\pi {x}^{-} 1\right)}{1}$

$= {\lim}_{x \rightarrow 2} \left(- \sin \left(\frac{\pi}{x}\right)\right) \left(- \pi {x}^{-} 2\right)$

$= {\lim}_{x \rightarrow 2} \frac{\pi \sin \left(\frac{\pi}{x}\right)}{x} ^ 2$

The limit can now be evaluated:

$= \frac{\pi \sin \left(\frac{\pi}{2}\right)}{2} ^ 2$

$= \frac{\pi}{4}$