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With this $\tan 2x$ on top, I can't really think of any ways to solve it.

$$ \lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}} $$

So is it solvable and how?

What I have came up so far:

$$ \begin{align} \lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}} \\ =\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{4x-2\pi} \end{align} $$

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Have you tried L'Hopital's Rule – Hawk Oct 4 '12 at 5:23
It's not true that $\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}} =\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{4x-2\pi}$. It is true that $\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}} =\lim_{x \to \frac{\pi}{2}} \frac{4\tan 2x}{4x-2\pi}$. If you multiply the denominator by something, you have to multiply the numerator by the same thing. – Michael Hardy Oct 4 '12 at 15:09
up vote 3 down vote accepted

Let $y=x-\frac{\pi}2$. Then $2x=2y+\pi$

$$\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}}=\lim_{y\to0}\frac{\tan(2y+\pi)}y=\lim_{y\to0}\frac{\tan 2y}y=2\lim_{y\to0}\frac{\tan 2y}{2y}\;.$$

That’s a limit that you should either know or be able to calculate fairly easily.

(Of course you can also use l’Hospital’s rule, which works easily on this limit.)

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Thanks.​​​​​​​​ – Derek 朕會功夫 Oct 4 '12 at 5:42
@Derek: My pleasure. – Brian M. Scott Oct 4 '12 at 5:43

Use L'Hôpital's rule. \begin{align} \lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}} &= \lim_{x \to \frac{\pi}{2}} 2\sec^2 2x \\ &= 2 \end{align}

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Thanks.​​​​​​​​ – Derek 朕會功夫 Oct 4 '12 at 5:41

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