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.
 A: 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
$$\frac{e^t-1}t=\frac{y}{\log (1+y)}\to 1$$
A: Your limit is equivalent to the well-known limit $\lim_{t\to 0}\frac{e^t-1}{t}=1$, but the most straightforward way to prove the last identity depends on your definition of the exponential function. For instance, if it is
$$ e^x = \sum_{n\geq 0}\frac{x^n}{n!} $$
there is nothing to prove. If it is $e^x=\lim_{n\to +\infty}\left(1+\frac{x}{n}\right)^n$ you still have that in a neighbourhood of the origin $e^x=1+x+o(x)$, by Bernoulli's inequality or AM-GM.
A: The question is how you have defined $e^x$. If it is as a power series, it is a bit silly to avoid Taylor series.
If you have defined the exponential as the inverse function of $\log x:=\int_1^x \frac1t dt$, then it is easy to show it is its own derivative and is $1$ at $0$
If you have defined $e^x=\lim_{n\to\infty}\left(1+\frac x n\right)^n$ you need to work slightly harder to show it is its own derivative and is $1$ at $0$, but it is not impossible.
A: MVT:
1) x \ge 0.
$e^x -1 = {\displaystyle \int_{0}^{x}} e^t dt = e^{y}(x-0)$, where
$y \in [0,x]$.
$\lim_{x \rightarrow 0} \dfrac{e^x-1}{x} =$ 
$\lim_{x \rightarrow 0} e^{y(x)} = e^{0}$,
since the exponential function  
is continuos and  $\lim_{x \rightarrow 0} y(x)=0$.
2) $x \le 0$.
$1- e^x ={\displaystyle \int_{x}^{0}} e^{t} dt  =$ 
$e^r (0-x)$, where $r \in [x,0].$
$\dfrac{e^x-1}{x}= e^r.$
Proceed to take the limit $x \rightarrow 0$,  as in case 1).
3) Right limit = Left limit, 
hence  $\lim_{x \rightarrow 0} \dfrac{e^x-1}{x}=1$.
A: If $e$ is defined as $$e^x=\sum_{i=0}^{\infty}\frac{x^i}{i!}$$ Then the rest is immidiate, since $$\frac{e^x-1}{x}=\sum_{i=0}^{\infty}\frac{x^i}{(i+1)!}=1+\frac{x}{2}+\dots$$
A: As you say f[x] = e^x[e^x-1][e^x+1]/x and
e^x[e^x -1]/x =1 as x tends towards 0 
So your remaining factor [e^x +1] as x tends towards 0 is
e^0+1 = 2
f[x] =1[2] =2 as x tends towards 0
