## Calculus (3rd Edition)

$2$
By substituting, we get the intermediate form $\frac{0}{0}$. So, by factorizing the numerator, we have $$\lim _{x \rightarrow 0} \frac{4^{3x}-4^x}{4^x-1}=\lim _{x \rightarrow 0} \frac{4^x(4^x+1)(4^x-1)}{4^x-1}\\ =\lim _{x \rightarrow 0} 4^x(4^x+1)=4^0(4^0+1)=2.$$