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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
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5 Answers 5

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
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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
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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
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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
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hint let y=2x then$$\lim_{x\to0}\frac{e^{2x}-1}{3x}=\frac23\lim_{2x\to0}\left(\frac{e^{2x}-1}{2x}\right)=\frac23\lim_{x\to0}\frac{e^{y}-1}{y}$$ then use : $\displaystyle\exp(y)=\sum_{k=o}^\infty\frac{y^k}{k!}$

$$e^y-1=\sum_{k=1}^\infty\frac{y^k}{k!}$$

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