$\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?

Answer 1

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.

Answer 2

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

Answer 3

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

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

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

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

Answer 4

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)

Answer 5

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 : $\displaystyle\exp(y)=\sum_{k=o}^\infty\frac{y^k}{k!}$

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