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How to find: $$\lim_{x \to \frac{\pi}{2}} \frac{\tan{2x}}{x - \frac{\pi}{2}}$$ I know that $\tan(2\theta)=\frac{2\tan\theta}{1-\tan^{2}\theta}$ but don't know how to apply it here.

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up vote 5 down vote accepted

Put $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$, you have $h \to 0$.

Then you have \begin{align*} \lim_{x \to \frac{\pi}{2}} \frac{\tan{2x}}{x-\frac{\pi}{2}} &= \lim_{h \to 0} \: \frac{\tan{2\bigl(\frac{\pi}{2}+h\bigr)}}{h} \\ &=\lim_{h \to 0} \: \frac{\tan(\pi + 2h)}{h} \\ &= \lim_{h \to 0} \: \frac{\tan(2h)}{h} \end{align*}

Can you do it from here?

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Yes, thanks. Now I just have to use the fact that $\lim_{h \to 0} \frac{\sin{2h}}{2h} = 1$ :) – Sura Sep 3 '11 at 20:42

l'Hospital Rule works here. The indetermination is $0/0$.

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thanks beni. Unfortunately, i don't have the rep to upvote :( – Sura Sep 3 '11 at 20:42
A correct and short solution, but I strongly object to using l'Hospital in the case of the denominator being a linear function: first, l'Hopital becomes an overkill. Moreover, the value at question is just the derivative of the numerator $tan(2x)$ at $\frac{\pi}{2}$. So I think this solution hides what's going on. – Ofir Sep 3 '11 at 21:21
@Ofir: Indeed, using l'Hospital is a bit of an overkill sometimes, but a faster solution is always welcome. For those who want to understand what's going on, it is always possible to calculate a limit avoiding l'Hospital. Thinking of what you said in your comment, one could just use the definition of the derivative to calculate the given limit, since in fact we need to calculate $\lim_{x \to \pi/2} \frac{f(x)-f(\pi/2)}{x-\pi/2}$ where $f(x)=\tan(2x)$. – Beni Bogosel Sep 4 '11 at 10:04

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