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.
Finding: $\lim_{x \to \frac{\pi}{2}} \frac{\tan{2x}}{x - \frac{\pi}{2}}$
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calculus
real-analysis
trigonometry
limits
2 Answers
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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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2Yes, thanks. Now I just have to use the fact that $\lim_{h \to 0} \frac{\sin{2h}}{2h} = 1$ :) – 2011-09-03
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l'Hospital Rule works here. The indetermination is $0/0$.
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0@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 $f(x)=\tan(2x)$. – 2011-09-04