limits - Evaluating the $lim_{xto0}frac{4^x-1}{8^x-1}$ without L'Hopital Rule

How to evaluate $\displaystyle \lim_{x\to0}\left(\frac{4^x-1}{8^x-1}\right)$ without L'Hopital rule?

When I evaluated this limit I got an indetermination, $\frac{0}{0}$. I learned that in a rational function when one get $\frac{0}{0}$ indeterminated form, one should look for the common terms between numerator and denominator by factoring. But I can't figure out how to find the common terms in this case. Can you help me? Thanks.

Answer

Note that

$$4^x-1=(2^2)^x-1=(2^x)^2-1=\color{red}{(2^x-1)}(2^x+1)$$
$$8^x-1=(2^3)^x-1=(2^x)^3-1=\color{red}{(2^x-1)}(2^{2x}+2^x+1)$$