Question on limit of power of rational functions Let $k$ be a fixed number and suppose that $q_k$ is a polynomial with rational coefficients. In other words $$q_k(y) = \sum_{i=0}^{k}b_i y^i$$ where $b_i, \quad 0 \le i \le k$ are rationals.
Here is my question: I need want to show that either $$\lim_{n \rightarrow \infty} \displaystyle \frac{2^n}{\displaystyle e^{q_k(n)}} = 0$$ or $$\lim_{n \rightarrow \infty} \displaystyle \frac{\displaystyle e^{q_k(n)}}{2^n} = 0.$$
I know that $$ \lim_{n \rightarrow \infty} \frac{1}{2^n} = 0$$ so the first thing that comes to my mind is that $$\lim_{n \rightarrow \infty} \displaystyle \frac{\displaystyle e^{q_k(n)}}{2^n}$$  has to be zero if the $\lim_{n \rightarrow \infty}\displaystyle e^{q_k(n)}$ is finite which I think is infinty. How can I approach this question? Any help on this will be great.
 A: There are three cases to consider, depending on whether the degree $k$ is $0$, $1$ or greater than $1$.  The only one that's not really obvious is $k=1$.  For this, 
 the important fact is that $\ln(2)$ is irrational.  This follows from the 
fact that $e$ is transcendental.
A: The proof is simple and its sketch is already provided in Robert Israel’s answer.
Decreasing $k$ if necessarily, without loss of generality we may assume $b_k\ne 0$. Put $x_n=\frac{2^n}{\displaystyle e^{q_k(n)}}=e^{n\ln 2-q_k(n)}$ and $y_n=n\ln 2-q_k(n)$.
So if $\lim_{n\to\infty} y_n=-\infty$ then  $\lim_{n\to\infty} x_n=\lim_{n\to\infty} e^{y_n}=0$. If $\lim_{n\to\infty} y_n=+\infty$ then $\lim_{n\to\infty} x_n^{-1}=\lim_{n\to\infty} e^{-y_n}=0$. Thus it suffices to show that $\lim_{n\to\infty} y_n=-\infty$ or $\lim_{n\to\infty} y_n=+\infty$. This clearly holds if $k>1$. If $k=0$ then $\lim_{n\to\infty} y_n=+\infty$, so  $\lim_{n\to\infty} x_n^{-1}=\lim_{n\to\infty} e^{-y_n}=0$. 
It remains to consider case $k=1$. Then $y_n=n(\ln 2-b_1)-b_0$. To show that $\lim_{n\to\infty} y_n=+\infty$ or $\lim_{n\to\infty} y_n=-\infty$ it suffices to remark that $\ln 2=\log_e 2$ is irrational, so $\ln 2-b_1\ne 0$. 
A: You know that $e \approx 2.7 > 2.5= \dfrac {5}{2}$.
You say that case $k=0$ is clear to you.
Let´s go on $k \geq 1$ then. 
Suppose that $b_k>0$. Then for large enough $n$ we will have $q_k(n) = \sum_{i=0}^{k}b_i n^i>n$. For large enough $n$ we will also have $e^{q_k(n)}> \left (\dfrac {5}{2} \right)^{q_k(n)}=\dfrac {5^{q_k(n)}}{2^{q_k(n)}}$.
So we have $0<\dfrac {1}{e^{q_k(n)}}<\dfrac {2^{q_k(n)}}{5^{q_k(n)}}$.
So we have $0<\dfrac {{2^{q_k(n)}}}{e^{q_k(n)}}<\dfrac {2^{q_k(n)}\cdot2^{q_k(n)}}{5^{q_k(n)}}=\dfrac {4^{q_k(n)}}{5^{q_k(n)}}=\left (\dfrac {4}{5} \right)^{q_k(n)}$.
Now take a break and eat your favourite sandwich and take the limit.
Can you now handle the case when $b_k<0$?
