# $\displaystyle \lim_{x\to 0} \frac{e^{2x}-1}{3x}$ help please?

So I have to find the

$$\displaystyle \lim_{x\to 0} \frac{e^{2x}-1}{3x}$$

I first solved this by L'hopital and got $\frac{2}{3}$ but now I read carefully and it says in my book that I shouldn't solve this by L'hopital..any hints?

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Are you allowed to use the derivative of $e^{2x}$ at $0$? – 1015 Feb 4 '13 at 18:20

A "fun" if not-so-edifying proof.

We know: $$\lim_{x\to 0}\frac {e^x-1}{x} = 1$$

And we have:

$$\lim_{x\to 0} \frac{e^x+1}{3} = \frac{2}{3}$$

Now multiply.

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Thomas,Y U SO SMART ? :( – rttrrt Feb 4 '13 at 18:38
I wouldn't necessarily call this a "smart" answer - I deliberately labeled it as "fun" because it is quick and not particularly edifying. The other answers are all more "general" in how they essentially apply L'Hopital without applying L'Hopital... – Thomas Andrews Feb 4 '13 at 18:50

Hint: The derivative of $e^{2x}$ at $0$ is $$\lim_{x\to 0}\frac{e^{2x}-e^{2\cdot 0}}{x-0}$$

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$$\lim_{x\to0}\frac{e^{2x}-1}{3x}=\frac23\lim_{2x\to0}\left(\frac{e^{2x}-1}{2x}\right)=\frac23$$

Alternatively,if we are allowed to use $$e^y=1+\frac y1+\frac{y^2}{2!}+\cdots$$

Then $$\lim_{x\to0}\frac{e^{2x}-1}{3x}$$ $$=\frac13\lim_{x\to0}\frac{1+2x+\frac{(2x)^2}{2!}+O(x^3)-1}x$$

$$=\frac13\lim_{x\to0}(2+O(x))=\frac23$$

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Did you really mean $\frac{3}{2}$? – Amzoti Feb 4 '13 at 18:27
@Amzoti, are you taking about the 1st method? – lab bhattacharjee Feb 4 '13 at 18:28
AH - that looks better :-)! Regards – Amzoti Feb 4 '13 at 18:29
@Amzoti, sorry for the horrible typo. Thanks for your observation. – lab bhattacharjee Feb 4 '13 at 18:31
Shouldn't it be $\Omega(x^3)$ instead of $O(x^3)$? – Paresh Feb 4 '13 at 18:36

Let $e^{2x}-1=y$ that yields

$$\displaystyle \lim_{y\to 0} \frac{2y}{3 \ln(1+y)}=\frac{2}{3}$$ because $\lim_{y\to0} (1+y)^{1/y}=e$

Sister. (It's the most elementary proof I know)

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I think it'd be better if you mention $\ln$ function is continuous at $e$, just my opinion. – julypraise Feb 4 '13 at 18:46