# Question c246f

Dec 9, 2017

Diverges, $\pm \infty$, Check below.

#### Explanation:

Really nice one!

Making your life easier would be to get rid of $\cos x$ and $\left(\frac{0}{0}\right)$

A good idea would be to set color(blue)x-color(blue)π/color(blue)2color(blue)=color(blue)u
The process of thinking behind this is to create cos(π/2+...) which equals to $- \sin \left(\ldots\right)$ which we can work with a lot easier!

lim_(xrarrπ/2)(1/cosx-1/(π-2x))

x-π/2=u
x->π/2
$u \to 0$

=lim_(urarr0)[1/cos(π/(2)+u)-1/(π-2(π/2+u))] $=$

=lim_(urarr0)(-1/sinu-1/(π-2*π/2-2u)) $=$

= lim_(urarr0)(-1/sinu-1/(π-π-2u)) $=$

$= {\lim}_{u \rightarrow 0} \left(- \frac{1}{\sin} u + \frac{1}{2 u}\right)$ $=$

$= {\lim}_{u \rightarrow 0} \left(\frac{1}{2 u} - \frac{1}{\sin} u\right)$ $=$

$= {\lim}_{u \rightarrow 0} \left(\frac{\sin u - 2 u}{2 u \cdot \sin u}\right)$

Now here is the interesting part. I am lead to $\left(\frac{0}{0}\right)$ once again. I can work with Rules De L' Hospital but i want to make the solution more "approachable" for everyone.

A thought now would be to create ${\lim}_{u \rightarrow 0} \sin \frac{u}{u}$ which i know equals to $1$ . As you can tell though we will have problem with the denominator.
Let's see

$u \to 0$ so that means $u \ne 0$

so we have ${\lim}_{u \rightarrow 0} \left(\frac{\sin \frac{u}{u} - \frac{2 u}{u}}{2 u \cdot \sin \frac{u}{u}}\right)$ $=$

$=$ ${\lim}_{u \rightarrow 0} \left(\frac{\sin \frac{u}{u} - 2}{2 \sin u}\right)$----> $\left(- \frac{1}{0}\right)$

=?#

Now we need to explore what happens near $0$ for $y = \sin x$

graph{sinx [-10, 10, -5, 5]}

• When $x \to {0}^{+}$ , $x > 0$ and $\sin x > 0$

then the $\lim$ leads to $- \left(\frac{1}{0} ^ +\right)$ so $- \left(+ \infty\right)$
, $- \infty$

• When $x \to {0}^{-}$ , $x < 0$ and $\sin x < 0$

then the $\lim$ leads to $- \left(\frac{1}{0} ^ -\right)$ so $- \left(- \infty\right)$
, $+ \infty$

Eventually , ${\lim}_{u \rightarrow 0} \left(\frac{\sin \frac{u}{u} - 2}{2 \sin u}\right)$
$= - \infty$ ,
when $x \to {0}^{+}$

and

$+ \infty$
when $x \to {0}^{-}$