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$lim _{x ightarrow rac{pi}{2}} rac{ an 2 x}{x- rac{pi}{2}}$

Question:

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}$

Solution:

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}$

At $x=\frac{\pi}{2}$, the value of the given function takes the form $\frac{0}{0}$

Now, put $x-\frac{\pi}{2}=y$ so that $x \rightarrow \frac{\pi}{2}, y \rightarrow 0$

$\therefore \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}=\lim _{y \rightarrow 0} \frac{\tan 2\left(y+\frac{\pi}{2}\right)}{y}$

$=\lim _{y \rightarrow 0} \frac{\tan (\pi+2 y)}{y}$

$=\lim _{y \rightarrow 0} \frac{\tan 2 y}{y} \quad[\tan (\pi+2 y)=\tan 2 y]$

$=\lim _{y \rightarrow 0} \frac{\sin 2 y}{y \cos 2 y}$

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

$=\left(\lim _{2 y \rightarrow 0} \frac{\sin 2 y}{2 y}\right) \times \lim _{y \rightarrow 0}\left(\frac{2}{\cos 2 y}\right) \quad[y \rightarrow 0 \Rightarrow 2 y \rightarrow 0]$

$=1 \times \frac{2}{\cos 0} \quad\left[\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$=1 \times \frac{2}{1}$

$=2$

 

 

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