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I'm studying the limits to $\lim_{x\rightarrow 0}\frac{e^{2x}-1}{\sin 3x}$. According to my text book, I should calculate it like this:

$$\frac{e^{2x}-1}{\sin 3x} = \frac{e^{2x}-1}{2x} \times \frac{3x}{\sin 3x} \times \frac{2}{3} \rightarrow 1 \times 1 \times \frac{2}{3}$$

While it makes sense, I don't understand the use of 2 and 3. Wouldn't any other numbers make as much sense? Why are these chosen and how?

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I think it's $e^{2x}$ in place of $e^2$ and use $$\lim_{h\to0}\frac{\sin h}h=1,\lim_{h\to0}\frac{e^h-1}h=1$$ –  lab bhattacharjee May 22 '13 at 16:21
    
Ah, yes, I meant 2x, question is updated. –  Quispiam May 22 '13 at 16:24
    
I think what lab said suffices. –  Parth Kohli May 22 '13 at 16:25
    
You know that $\lim_{x\to0}(\sin x)/x=1$, so $\lim_{x\to0}(\sin 3x)/(3x)=1$. –  egreg May 22 '13 at 16:29

2 Answers 2

Easier, use Taylor series expansion around 0: $e^{2x} =1+2x+2x^2 +O(x^3), \ \sin 3x = 3x-\frac{9x^3}{2} + O(x^5)$. Plug in this expansion into your expression, do a bit of algebra and the result easily comes out.

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L'Hopital's Rule might be an earlier covered topic than Taylor series expansion. –  hardmath May 22 '13 at 17:08
    
@hardmath: you are probably right. Just this approach seems to be the easiest. –  Alex May 22 '13 at 17:10
    
I'm not a huge fan of l'Hopital's Rule. Power series, if you understand them, certainly clear up the mystery that initially surrounds l'Hopital. –  hardmath May 22 '13 at 18:27

The fractions involving $e$ and $\sin$ turn out to be the very expressions (for $\sin$: reciprocal) that you use to calculate the derivatives of $\exp$ and $\sin$ at $0$. (This amounts to what lab said in a comment, but maybe show sth emotivation behind it)

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