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