# How do you find the limit of [(pi)tan(x)-(2x)/(cos(x))] as x approaches pi/2?

Jun 3, 2016

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

#### Explanation:

$\pi \tan \left(x\right) - \frac{2 x}{\cos} \left(x\right) \equiv \frac{\pi \sin \left(x\right) - 2 x}{\cos} \left(x\right)$

Calculating the series expansion of
$\pi \sin \left(x\right) - 2 x$ and $\cos \left(x\right)$ in the vicinity of $\frac{\pi}{2}$ we get

$\pi \sin \left(x\right) - 2 x = - 2 \left(x - \frac{\pi}{2}\right) - \frac{1}{2} \pi {\left(x - \frac{\pi}{2}\right)}^{2} + \cdots$
$\cos \left(x\right) = - \left(x - \frac{\pi}{2}\right) + \frac{1}{6} {\left(x - \frac{\pi}{2}\right)}^{3} + \cdots$

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

Jun 3, 2016

2

#### Explanation:

$L H S = {\lim}_{x \to \frac{\pi}{2}} \left[\pi \tan \left(x\right) - \frac{2 x}{\cos} \left(x\right)\right]$

Let $\text{ } x = \frac{\pi}{2} - z$

$\therefore x \to \frac{\pi}{2} \implies z \to 0$

Hence the given expression becomes

${\lim}_{z \to 0} \left[\pi \tan \left(\frac{\pi}{2} - z\right) - \frac{2 \left(\frac{\pi}{2} - z\right)}{\cos} \left(\frac{\pi}{2} - z\right)\right]$

=lim_{z->0}[pi cot(z))-(pi-2z)/sin(z)]

$= {\lim}_{z \to 0} \left[\frac{\pi \cos \left(z\right)}{\sin} \left(z\right) - \frac{\pi}{\sin} \left(z\right) + 2 \frac{z}{\sin} \left(z\right)\right]$

$= {\lim}_{z \to 0} \left[\frac{\pi}{\sin} \left(z\right) \left(\cos \left(z\right) - 1\right) + 2 \frac{z}{\sin} \left(z\right)\right]$

$= {\lim}_{z \to 0} \left[- \frac{2 \pi {\sin}^{2} \left(\frac{z}{2}\right)}{2 \sin \left(\frac{z}{2}\right) \cos \left(\frac{z}{2}\right)} + 2 \frac{z}{\sin} \left(z\right)\right]$

$= {\lim}_{z \to 0} \left[- \pi \tan \left(\frac{z}{2}\right) + 2 \frac{z}{\sin} \left(z\right)\right]$

$= \left[- \pi {\lim}_{z \to 0} \tan \left(\frac{z}{2}\right) + 2 {\lim}_{z \to 0} \frac{z}{\sin} \left(z\right)\right]$

$= - \pi \times 0 + 2 \times 1 = 2$