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

Question

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?

Are you allowed to use the derivative of $e^{2x}$ at $0$? – Julien Feb 04 '13 at 18:20 — Julien, Feb 04 '13 at 18:20

Thomas Andrews · Answer 1 · 2013-02-04T19:39:47.280

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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.

edited Feb 04 '13 at 19:39

answered Feb 04 '13 at 18:31

Thomas Andrews

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Thomas,Y U SO SMART ? :( – rttrrt Feb 04 '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 04 '13 at 18:50

score 3 · Answer 2 · answered Feb 04 '13 at 18:19

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

answered Feb 04 '13 at 18:19

Nameless

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lab bhattacharjee · Answer 3 · 2013-02-04T18:27:11.817

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

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

edited Feb 04 '13 at 18:27

answered Feb 04 '13 at 18:20

lab bhattacharjee

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Did you really mean $\frac{3}{2}$? – Amzoti Feb 04 '13 at 18:27
@Amzoti, are you taking about the 1st method? – lab bhattacharjee Feb 04 '13 at 18:28
AH - that looks better :-)! Regards – Amzoti Feb 04 '13 at 18:29
@Amzoti, sorry for the horrible typo. Thanks for your observation. – lab bhattacharjee Feb 04 '13 at 18:31
Shouldn't it be $\Omega(x^3)$ instead of $O(x^3)$? – Paresh Feb 04 '13 at 18:36
@labbhattacharjee: you are welcome and no problem, my fat fingers are always getting me into trouble! Regards – Amzoti Feb 04 '13 at 18:38
@Paresh, http://en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics – lab bhattacharjee Feb 04 '13 at 18:39
I see! Thanks for that. – Paresh Feb 04 '13 at 18:42
Paresh is right but all $\Omega, \Theta, O$ are also simply written as $O$, – Feb 04 '13 at 18:46

score 2 · Answer 4 · answered Feb 04 '13 at 18:41

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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)

answered Feb 04 '13 at 18:41

user 1591719

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I think it'd be better if you mention $\ln$ function is continuous at $e$, just my opinion. – le4m Feb 04 '13 at 18:46

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

4 Answers4