How can I show $\lim_{x\to 0}\frac{e^{3x}-e^x}{x}=2$ without derivatives? So I was looking through a Calculus book, in their first chapter section 2 about limits and I came across the following problem: $$ \lim_{x\to 0} \dfrac{e^{3x}-e^x}{x}=2 $$ Immediately, I took the limit using L'Hopital's, but this is only the first chapter and they have yet to cover derivatives. Other than using Taylor Series (which would be taught later as well) is there another way to answer this question? I had thought of some factorization $$\dfrac{e^{3x}-e^x}{x}=\dfrac{e^x(e^{2x}-1)}{x}=\dfrac{e^x(e^x-1)(e^x+1)}{x}$$ Which made me think the problem could devolve into me showing that: $$ \lim_{x\to 0} \dfrac{e^x(e^x-1)}{x}=1 $$ and similarly, $$ \lim_{x\to 0} \dfrac{(e^x-1)}{x}=1. $$ However, even here I tend to think of Taylor series to prove this. Any help is immensely appreciated.

Given the standard limit as $t \to 0$ $\frac{e^t-1}t \to 1$ we have that

$$\dfrac{e^{3x}-e^x}{x}=2e^x \frac{e^{2x}-1}{2x}\to2\cdot1\cdot1$$

To prove the standard limit recall that

$$e=\lim_{y\to \infty}\left(1+\frac1y\right)^y \implies 1=\log e=\lim_{y\to \infty}y\log \left(1+\frac1y\right) \implies \lim_{x\to 0} \frac{\log(1+x)}{x} \to 1 $$

therefore by $y=e^t-1 \to 0 \implies t=\log (1+y)$ we have

$$\frac{e^t-1}t=\frac{y}{\log (1+y)}\to 1$$