Giải thích các bước giải:
$\lim\dfrac{2^2+3^n-4^n}{2^n+3^{n+1}+4^n+1}$
$=\lim\dfrac{4+3^n-4^n}{2^n+3.3^{n}+4^n+1}$
$=\lim\dfrac{\dfrac{4}{4^n}+(\dfrac{3}{4})^n-1}{(\dfrac{2}{4})^n+3.(\dfrac{3}{4})^{n}+1+\dfrac{1}{4^n}}$
$=\dfrac{0+0-1}{0+3.0+1+0}=-1$
