How do I find this limit? $\lim_{n\to\infty}\frac{(n+1)^2}{4^n}$ $$\lim_{n\to\infty}\frac{(n+1)^2}{4^n}$$ The answer says zero is the limit, how this is I do not understand, I've tried H'opitals rule , didn;t work out for me, anyone have any suggestions on how to solve this limit ?
By l'Hospital's rule, you have
$$
\lim_{n\to\infty}\frac{(n+1)^2}{4^n}=\lim_{n\to\infty}\frac{2(n+1)}{4^n\cdot\log 4}=\lim_{n\to\infty}\frac{2}{4^n(\log 4)^2}=\frac{2}{(\log 4)^2}\lim_{n\to\infty}\frac{1}{4^n}=0.
$$
$4^n$ increases much faster than $(n+1)^2$ and when $n$ tends to infinity, the denominator will be very large compared to the numerator. Hence the limit is $0$.