# Question #57cc0

Jan 23, 2018

${\lim}_{x \to \frac{\pi}{2}} \left(\frac{\pi}{2} - x\right) \tan \left(x\right) \to 0 \times \infty$

Rewrite the equation as:

${\lim}_{x \to \frac{\pi}{2}} \left(\frac{\pi}{2} - x\right) \tan \left(x\right) = {\lim}_{x \to \frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - x\right) \sin \left(x\right)}{\cos} \left(x\right) \to \frac{0}{0}$

This indeterminate form can be evaluated using L'Hopital's rule.

The derivative of the numerator will be:

$\frac{d}{\mathrm{dx}} \left(\frac{\pi}{2} - x\right) \sin \left(x\right) = - \sin \left(x\right) + \left(\frac{\pi}{2} - x\right) \cos \left(x\right)$

(using the product rule).

The derivative of the denominator:

$\frac{d}{\mathrm{dx}} \cos \left(x\right) = - \sin \left(x\right)$

So:

${\lim}_{x \to \frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - x\right) \sin \left(x\right)}{\cos} \left(x\right)$
$= {\lim}_{x \to \frac{\pi}{2}} \frac{- \sin \left(x\right) + \left(\frac{\pi}{2} - x\right) \cos \left(x\right)}{- \sin \left(x\right)} = \frac{- 1}{-} 1 = 1$