# How do you find the limit of  (x−pi/4)tan(2x)  as x approaches pi/4?

May 26, 2016

Please see the explanation section, below.

#### Explanation:

lim_(xrarrpi/4) (x−pi/4)tan(2x)

I'd like to use the fundamental trigonometric limts, so I'll change the variable.

Let $u = x - \frac{\pi}{4}$,

so that $x = u + \frac{\pi}{4}$ and $x \rightarrow \frac{\pi}{4}$ is equivalent to $u \rightarrow 0$.

Nww we have been asked to find

${\lim}_{u \rightarrow 0} u \tan 2 \left(u + \frac{\pi}{4}\right) = {\lim}_{u \rightarrow 0} u \tan \left(2 u + \frac{\pi}{2}\right)$

Rewriting $\tan$ as $\frac{\sin}{\cos}$, we get

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

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

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

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

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

$= \left(- \frac{1}{2}\right) \left(1\right) \left(1\right) = - \frac{1}{2}$

lim_(xrarrpi/4) (x−pi/4)tan(2x) = -1/2

We have that

lim_(xrarrpi/4) (x−pi/4)tan(2x)= lim_(xrarrpi/4) (x-pi/4)*sin(2x)/(cos2x)

Hence the limit is the form $\frac{0}{0}$ we apply L'Hopital Law hence

lim_(xrarrpi/4) (d[(x-pi/4)*sin2x]/dx)/(d(cos2x)/dx)= lim_(xrarrpi/4) [sin2x+2*(x-pi/4)cos2x]/(-2sin2x)=-1/2