# How do you find the limit of  (x-pi)/(sinx) as x approaches pi?

May 5, 2018

${\lim}_{x \rightarrow \pi} \frac{x - \pi}{\sin x} = {\lim}_{x \rightarrow \pi} \frac{1}{\cos} x = \frac{1}{-} 1 = - 1$

#### Explanation:

show the steps

${\lim}_{x \rightarrow \pi} \frac{x - \pi}{\sin x}$ Direct compensation product equal $\left(\frac{0}{0}\right)$

we must use L'Hopital's Rule

${\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ if the direct compensation product equal $\left(\frac{0}{0}\right)$

$f \left(x\right) = x - \pi$
$f ' \left(x\right) = 1$
$g \left(x\right) = \sin x$
$g ' \left(x\right) = \cos x$
${\lim}_{x \rightarrow \pi} \frac{x - \pi}{\sin x} = {\lim}_{x \rightarrow \pi} \frac{1}{\cos} x = \frac{1}{-} 1 = - 1$