Solve $\lim_\limits{x\to 0}\frac {e^{3x}-1}{e^{x}-1} $ I have problem with  $$\lim_{x\to 0}\frac {e^{3x}-1}{e^{x}-1} $$
I have no idea what to do first. 
 A: HINT: Note that $x^3-1=(x-1)(x^2+x+1)$
A: Hint:
Use equivalents:
$$\mathrm e^{ax}-1\sim_0 ax,\quad\text{hence}\quad \frac{\mathrm e^{ax}-1}{\mathrm e^x-1}\sim_0 \frac{ax}x=a.$$
Alternative hint:
$$\frac{\mathrm e^{ax}-1}{x}\xrightarrow[x\to0]{}(\mathrm e^{ax})'\,\Big\lvert_{x=0}$$
A: Let's apply L'Hospital's Rule here; i.e.,
$$\lim_{x\to 0}\frac {e^{3x}-1}{e^{x}-1}= \lim_{x\to 0}\frac {3e^{3x}}{e^{x}}=\bbox[5px,border:2px solid #F0A]3\,.$$
OR We can try substitution if you're not familiar with L'H rule:
$$t=e^x\Rightarrow \color{blue}{t\rightarrow1 \: as \: x\rightarrow0} \: \Rightarrow\lim_{x\to 0}\frac {e^{3x}-1}{e^{x}-1}=\lim_{t\to 1}\frac {t^3-1}{t-1}=\lim_{t\to 1}\frac {(t-1)(t^2+t+1)}{t-1}=\lim_{t\to 1}(t^2+t+1)=\bbox[5px,border:2px solid #F0A]3$$

What you see above is the graph of $y=\frac{t^3-1}{t-1}$. As you can see it is not defined at $t=1$, but it has a limit that is equal to $3$ at $t=1$.
A: My first thought was to write it as
$$
\frac{\left(\lim\limits_{x\to 0} \dfrac{e^{3x}-e^{3\,\cdot\,0}}{x-0} \right)}{\left(\lim\limits_{x\to 0} \dfrac{e^x-e^0}{x-0} \right)} = \frac{\left.\dfrac d {dx} e^{3x} \,\right|_{x=0}}{\left.\dfrac d {dx} e^x \,\right|_{x=0}} = \text{etc.}
$$
But the simpler way is to write $e^{3x}-1 = (e^x-1)(e^{2x}+e^x+1)$ and then cancel.
A: without using L'Hopital's rule
$$\lim_{x\to 0}\frac{e^{3x}-1}{e^x-1}=\lim_{x\to 0}\frac{(e^x-1)(e^{2x}+e^x+1)}{e^x-1}=\lim_{x\to 0} e^{2x}+e^x+1=3$$
A: Hint Recognize $e^{3x}-1$ as a difference of cubes.
A: Looking at Taylor series for $x \to 0$:
$$
\lim_{x \to 0} \frac{e^{3x}-1}{e^x-1} = \frac{1 + 3x - 1 + \mathcal{O}(x^2)}{1 + x - 1 + \mathcal{O}(x^2)} = \frac{3x + \mathcal{O}(x^2)}{x + \mathcal{O}(x^2)} = 3 
$$
