$\lim_{ n\to \infty} \frac{n^{100}}{1.01^n} = ?$ 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?
 A: The good way to find this is to have some experience with asymptotic growth rates and know that any exponential function eventually outgrows any polynomial (if the exponential function is increasing at all).
This immediately tells us that your limit is $0$.
If you don't have this knowledge already, exercises like this are exactly designed to give it to you. A reasonably painless way to see the result is to take the logarithm:
$$ \log\frac{n^{100}}{1.01^n} = 100 \log n - n \log(1.01) $$
The slope of $100\log n$ is $\frac{100}{n}$, which goes to $0$. So eventually $n$ will get large enough that the slope of $n\mapsto 100\log n$ is smaller than, say, half of $\log(1.01)$, and then it can only be a matter of time before the logarithm gets (and stays) below any desired negative value.
Since the logarithm of $\frac{n^{100}}{1.01^n}$ goes towards $-\infty$, the fraction itself goes to $0$.

Note that on its way to $0$, the fraction $\frac{n^{100}}{1.01^n}$ does reach values as high as about $10^{356}$, around $n=10^4$.
A: Write $n=1.01^{\log_{1.01}n}$ to then get the reformulation
$$\lim_{n\to\infty}\frac {1.01^{100\log_{1.01}n} }{1.01^n}=\lim_{n\to\infty}1.01^{100\log_{1.01}(n)-n}=1.01^{\lim_{n\to\infty}(100\log_{1.01}(n)-n)}=0$$
This simply follows from $n $ "increasing faster" than $100\log_{1.01}(n)$ for large $n $ (the difference of their derivatives tends to infinity) making $\lim_{n\to\infty}(100\log_{1.01}(n)-n)=-\infty $.
A: If you know about series, then apply the ratio test to $\displaystyle \sum \frac{n^{100}}{1.01^n}$ :
$$
\frac{\dfrac{(n+1)^{100}}{1.01^{n+1}}}
{\dfrac{n^{100}}{1.01^n}}
=\frac{(1+\frac1n)^{100}}
{1.01}
\to 
\frac{1}{1.01}
< 1
$$
This means that the series converges, and so 
$$
\frac{n^{100}}{1.01^n} \to 0
$$
A: One way to argue that is not quite a valid proof, but gives you an idea of what the limit ought to be (and is, in my opinion, the easiest one to do in your head, without writing anything down) is the following: What happens when you double $n$? Well, the numerator is increased by a factor of $2^{100}$, which is quite a lot. However, the denominator is squared. To appreciate what that means, go so far out in the sequence that $1.01^n$ is larger than $2^{100}$. It happens quite close to $n = 7000$ (between $n = 6966$ and $n = 6967$, to be specific), so let's say that's where we are.
Now, going from $a_{7000} = \frac{7000^{100}}{1.01^{7000}}$ to $a_{14\,000} = \frac{14\,000^{100}}{1.01^{14\,000}}$ means, as I said above, that the numerator is multiplied by $2^{100}$ and the denominator is multiplied by itself, which is about $2^{100}$ as well (and becomes close to $2^{200}$). That means that these two terms of the sequence are approximately equal. What happens if we double $n$ again, to $28\,000$? The numerator is still multiplied by $2^{100}$, but the denominator is squared, which means that it's multiplied by (about) $2^{200}$. So the value of the fraction decreases by a factor of $2^{100}$. The next doubling of $n$ will decrease the value of the fraction by a factor of $2^{700}$, and so on.
It's pretty clear that this goes toward $0$ as we continue. In order to make it a full proof you have to show that all the terms that we don't talk about go toward $0$ along with the ones we do study. That's a bit more work, and means that this approach probably isn't the best for a full proof.
A: 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 $n$.
