$\lim_{n \to \infty}\frac{n^{100}}{2^n+n^2+5}$ $$\lim_{n \to \infty}\frac{n^{100}}{2^n+n^2+5}$$
I do not know how to handle $n^{100}$ I thought that $2^n$>$n^{100}$ and therefore the 
$$0=\lim_{n \to \infty}\frac{n^{100}}{3\cdot2^n}\leq\lim_{n \to \infty}\frac{n^{100}}{2^n+n^2+5}\leq\lim_{n \to \infty}\frac{n^{100}}{2^n}=0$$
$$\lim_{n \to \infty}\frac{n^{100}}{2^n+n^2+5}=0$$
Am I right?
 A: You may write, as $n$ is great:
$$
\frac{n^{100}}{2^n+n^2+5}=\frac{\frac{n^{100}}{2^n}}{1+\frac{n^{2}}{2^n}+\frac{5}{2^n}} \longrightarrow \frac01=0
$$ since you know that
$$\lim_{n \to \infty}\frac{n^{100}}{2^n}=\lim_{n \to \infty}\frac{n^{2}}{2^n}=\lim_{n \to \infty}\frac{5}{2^n}=0.$$
A: You say "I thought that $2^n>n^{100}$" and this is true for all sufficiently large $n$, however for small $n$ it is not true at all. 
Take $n=10$ for example. Then $2^{10} = 1024$, while $10^{100}$ is a $1$ with followed by $100$ zeros.
You are correct that the limit is $0$, but for small $n$ the value is huge, though not infinite. This is not in contradiction; indeed, it is precisely what this exercise should make your realize.   
A: here is a way to show $\lim_{n \to \infty}{n^{100} \over 2^n} = 0$ accepting that the limit exists.
we will use the fact that if $\lim_{n \to \infty}f(n) = L,$ then so is  $\lim_{n \to \infty}f(2n) = L,$
suppose $$L = \lim_{n \to \infty} {n^{100} \over 2^n} = \lim_{n \to \infty}{(2n)^{100} \over 2^{2n}} = \lim_{n \to \infty} {2^{100} \over 2^n} 2^{100-n} = L\lim_{n \to \infty} 2^{100 -n} = L \times 0 = 0$$
A: Write $100^n = 2^{100 \log_2 n}$. Then ${100^n \over 2^n} = 2^{100 \log_2 n -n}$. Since $\lim_{n \to \infty} (100 \log_2 n -n) = -\infty$, we have $\lim_{n \to \infty}{100^n \over 2^n}  = 0$.
To see why $\lim_{n \to \infty} (100 \log_2 n -n) = -\infty$, first look at
$\ln x - \alpha x$ for some $\alpha>0$. We have
$\ln x - \alpha x = \ln x - \alpha (x-1) - \alpha = \int_1^x ({1 \over t}-\alpha) dt  - \alpha$. If we choose $x_0 = { 2 \over \alpha}$ 
and let $C= \int_{x_0}^{x} ({1 \over t}-\alpha) dt - \alpha$, we have
$\ln x - \alpha x =  C + \int_{x_0}^{x} ({1 \over t}-\alpha) dt \le C - { \alpha \over 2} (x-x_0)$, from which we see that
$\lim_{x \to \infty} (\ln x - \alpha x) = -\infty$. Since $\log_2 x= {1 \over \ln 2} \ln x$, choosing $\alpha = { \ln 2 \over 100}$ gives the desired result.
