$\lim_{ n\to \infty} \frac{n^{100}}{1.01^n} = ?$ And how to prove it？ There are many such similar limits. But it seems that each of the proof is isolated, there are any good ways to solve it?

In all generality,
$$\lim_{n\to\infty}n^ab^n=\lim_{n\to\infty}\left(nb^{n/a}\right)^a=\lim_{n\to\infty}\left(nc^n\right)^a=\left(\lim_{n\to\infty}nc^n\right)^a,$$ if it exists, with $c:=\sqrt[a]b$.
Then every time you increment $n$, the expression between the parenthesis is multiplied by $$\frac{n+1}nc=\left(1+\frac1n\right)c.$$
The first factor gets closer and closer to $1$ so that in the end $c$ makes the decision.
If $c>1$, the factor is larger than $1$ and the limit goes to infinity.
If $c<1$, the factor will end up being smaller than $1$ and the limit goes to zero.
Then if $c=1$, the limit is just that of $n$.
