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22. Evaluate the following limits \lim_{x \rightarrow \pi /2 } \frac{\tan 2x }{x - \pi /2 }

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\lim_{x \rightarrow \pi /2 } \frac{\tan 2x }{x - \pi /2 }

The function takes zero-by-zero form when the limit is put directly, so we simplify the function and then put the limits.

Let's put

 y=x-\frac{\pi}{2}

Since we are changing the variable, limit will also change.

as

 x\rightarrow \frac{\pi}{2},y=x-\frac{\pi}{2}\rightarrow \frac{\pi}{2}-\frac{\pi}{2}=0

So, function in new variable becomes,

\lim_{y \rightarrow 0 } \frac{\tan 2(y+\frac{\pi}{2}) }{y+\pi/2- \pi /2 }

=\lim_{y \rightarrow 0 } \frac{\tan (2y+\pi) }{y }

As we know, this property   tan(\pi+x)=tanx

=\lim_{y \rightarrow 0 } \frac{\tan (2y) }{y }

=\lim_{y \rightarrow 0 } \frac{sin2y}{2y}\cdot\frac{2}{cos2y}

=1\times 2

=2   (Answer)

Posted by

Pankaj Sanodiya

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