Question

Evaluate the Given limit: $\lim_{x\rightarrow \frac{\pi}{2}}\frac{tan\,2x}{x}-\frac{\pi}{2}$

Answer 1

$\lim_{x\rightarrow \frac{\pi}{2}}\frac{tan\,2x}{x}-\frac{\pi}{2}$
At x =$\frac{\pi}{2}$, the value of the given function takes the form 0/0.
Now, put x - $\frac{\pi}{2}$ = y so that x\rightarrow$$\frac{\pi}{2} , y$\rightarrow$0
∴ $\lim_{x\rightarrow \frac{\pi}{2}}\frac{tan\,2x}{x}-\frac{\pi}{2}$ = $\lim_{y\rightarrow 0}\frac{tan\,2(y+\frac{\pi}{2})}{y}$
= $\lim_{y\rightarrow 0}\frac{tan(\pi+2y)}{y}$
= $\lim_{y\rightarrow 0}\frac{tan(2y)}{y}$ [ tan($\pi$+2y) = tan2y]
= $\lim_{y\rightarrow 0}$ $\frac{sin\,2y}{y\,cos\,2y}$
=$\lim_{y\rightarrow 0}$ ($\frac{sin\,2y}{2y}\times\frac{2}{cos\,2y}$)
=( $\lim_{2y\rightarrow 0}\frac{sin\,2y}{y}\times\lim_{y\rightarrow 0}\frac{2}{cos2y}$)[ y$\rightarrow$0$\Rightarrow$2y$\rightarrow$0]
=$1\times \frac{2}{cos\,0}$ [ $\lim_{x\rightarrow 0}$ $\frac{sin\,x}{x}$ = 1]
=$1\times\frac{2}{1}$
=2