# Question 16e49

Nov 21, 2017

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

#### Explanation:

Evaluate:

lim_(x->pi/2) cosx/(ln(x-pi/2+1)#

substituting $y = x - \frac{\pi}{2}$. Then for $x \to \frac{\pi}{2}$ we have $y \to 0$ and:

$\cos x = \cos \left(x - \frac{\pi}{2} + \frac{\pi}{2}\right) = \cos \left(y + \frac{\pi}{2}\right)$

Using the formula for the cosine of the sum of two angles:

$\cos \left(y + \frac{\pi}{2}\right) = \cos y \cos \left(\frac{\pi}{2}\right) - \sin y \sin \left(\frac{\pi}{2}\right) = - \sin y$

so:

${\lim}_{x \to \frac{\pi}{2}} \cos \frac{x}{\ln} \left(x - \frac{\pi}{2} + 1\right) = {\lim}_{y \to 0} - \sin \frac{y}{\ln} \left(1 + y\right)$

Divide numerator and denominator by y:

${\lim}_{x \to \frac{\pi}{2}} \cos \frac{x}{\ln} \left(x - \frac{\pi}{2} + 1\right) = {\lim}_{y \to 0} \frac{- \sin \frac{y}{y}}{\ln \frac{1 + y}{y}}$

and using the well known limits:

${\lim}_{t \to 0} \sin \frac{t}{t} = 1$

${\lim}_{t \to 0} \ln \frac{1 + t}{t} = 1$

we can conclude:

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