How to find: \lim_{x \to \frac{\pi}{2}} \frac{\tan{2x}}{x - \frac{\pi}{2}}

Question

How to find:

lim_{x \to \frac{π}{2}} \frac{\tan {2 x}}{x - \frac{π}{2}}

Answer 1

Put $x = \frac{π}{2} + h$ . As $x \to \frac{π}{2}$ , you have $h \to 0$
Then you have
$lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} = lim_{h \to 0} \frac{\tan 2 (\frac{π}{2} + h)}{h}$
$= \lim_{h \to 0} \frac{\tan (π + 2 h))}{h}$
$= lim_{h \to 0} \frac{\tan (2 h)}{h}$
Can you do it from here?

How to find: \lim_{x \to \frac{\pi}{2}} \frac{\tan{2x}}{x - \frac{\pi}{2}}

Answers (1)

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