Question: How do I prove that $\lim\_{n\to\infty}\frac{n^c}{b^n}=0$, where $c\in\mathbb{R}$ and $b>1$?
I've managed to prove it using L'Hospital rule working with $\frac{x^c}{b^x}$, but is there a simpler way using only sequences?

Answer: 
Another way to see it:
$$
\sum\_{n=1}^\infty \frac{n^c}{b^n} < \infty
$$
by the [ratio test](https://en.wikipedia.org/wiki/Ratio_test), hence the sequence must converge to $0$.

**edit:**
$$
\left| \frac{\frac{(n+1)^c}{b^{n+1}}}{\frac{n^c}{b^n}} \right| = \frac{(n+1)^c}{b^{n+1}} \cdot \frac{b^n}{n^c} = \left(\frac{n+1}{n}\right)^c \cdot \frac 1 b \to \frac{1}{b} < 1
$$
as $n\to\infty$.
Let $m$ be any integer greater than $c$ and $0$. Since $b-1>0$ we have for $n\geq m$: $$b^n=(b-1+1)^n=\sum\_{k=0}^n\binom{n}{k}(b-1)^k\geq\binom{n}{m}(b-1)^m=:g(n)$$
Now the right side $g(n)$ is a polynomial in $n$ of degree $m>c$ in $n$. It follows that $n^c/g(n)\to0$ as $n\to\infty$ and therefore by the squeeze theorem: $$\lim\_{n\to\infty}\frac{n^c}{b^n}=0$$