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. 

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 $t=1$.
