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. 

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.
