$$\lim \frac{2^n}{n^{100}}$$ as n goes to infinity of course.
I know that the form os $\frac{a_{n+1}}{a_n}$
$$\frac {\frac{2^{n+1}}{(n+1)^{100}}}{\frac{2^n}{n^{100}}}$$
$$\frac{2n^{100}}{(n+1)^{100}}$$
I am not clever enough to evaluate that limit. To me it looks like it should go to zero using the logic that exponents increase much more quickly than twice. Also I feel like I could reduce it down to $\frac{2}{n}$ but that gives an incorrect answer.
Marty has answered the problem of finding the original limit, which as Ted pointed out does not require the ratio test. However, in some circumstances, including this one, the ratio test does give a way to find the limit. So let's first look at your question of finding the limit of the ratio, then come back the original sequence.
As has been pointed out, it helps to note that $\dfrac{n^{100}}{(n+1)^{100}}=\left(\dfrac{n}{n+1}\right)^{100}$. It is easy to find the limit of $\dfrac{n}{n+1}$, e.g. by first writing it as $1-\dfrac{1}{n+1}$ or as $\dfrac{1}{1+\frac{1}{n}}$. Then the identity $\lim\limits_{n\to\infty} x_n^{100}=(\lim\limits_{n\to\infty} x_n)^{100}$ applies.
You should thus be able to conclude that $\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}=2$, if $a_n=\dfrac{2^n}{n^{100}}$. This implies that $\lim\limits_{n\to\infty}a_n=+\infty$. Informally, the limit of the ratios being $2$ means that when $n$ is large, $a_{n+1}$ is about twice as large as $a_n$, and then $a_{n+m}\approx 2^ma_n\to\infty$ as $m\to\infty$. More formally, it implies that the sequence is eventually monotone and therefore converges to a positive real number or $+\infty$. The finite case can be ruled out by contradiction, because $\lim\limits_{n\to\infty}{a_n}=L<+\infty$ would imply that $\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}=\dfrac{L}{L}=1.$ All that was required is that the limit of the ratio was greater than $1$. (If the limit of the ratio were less than $1$, the sequence would converge to $0$, and if the limit were $1$ the test would be inconclusive.)
In Prove that $n^k < 2^n$ for all large enough $n$ I showed that if $n$ and $k$ are integers and $k \ge 2$ and $n \ge k^2+1$, then $2^n > n^k$.
Set $k = 101$. Then, for $n \ge 101^2+1 = 10202$, $2^n > n^{101}$ or $\dfrac{2^n}{n^{100}} > n$ so $\lim_{n \to \infty} \dfrac{2^n}{n^{100}} = \infty$.
This obviously shows that $\lim_{n \to \infty} \dfrac{2^n}{n^k} = \infty$ for any positive integer $k$.
Marty and Jonas have done a great job of explaining the main idea of solving this problem, I just want to add to what Jonas mentioned about being able to find that $\dfrac{a_{n+1}}{a_n}$.
When expanding binomials, use the binomial theorem:
$$(a + b)^n = a^n + {n\choose 1} a^{n-1}b + {n\choose 2}a^{n-2}b^2 + \cdots + b^n.$$
Using this we can change the fraction that you had to:
$$\dfrac{2n^{100}}{n^{100} + 100n^{99} + 4950n^{98} + \cdots}.$$
If you notice on the bottom we have a polynomial of degree $100$, this shares the same degree of the monomial on the top, so we can disregard the other $100$ terms after $n^{100}$. This is why the limit as $n$ goes to infinity of $\dfrac{a_{n+1}}{a_n} = 2$ for our ratio test. Because the coefficient of the highest powered term on the numerator (2) divided by the coefficient of the highest power term in the denominator (1) is equal to 2.
I come here with my fancy way: look at the series
$$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{n^{100}}{2^n}$$
Since this is a positive series we can apply the quotient rule, say:
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{100}}{2^{n+1}}\frac{2^n}{n^{100}}=\frac12\left(\frac{n+1}n\right)^{100}\xrightarrow[n\to\infty]{}\frac12<1$$
The above means this positive series converges, and this means
$$\;\displaystyle{\frac{n^{100}}{2^n}\xrightarrow[n\to\infty]{}0\implies\frac{2^n}{n^{100}}\xrightarrow[n\to\infty]{}\infty}\;$$
