Prove that $\lim_{n \to \infty} k^nn^p= 0$ where $|k| < 1$ and $p>0$ Hello I am working through some problems in a book and came across this question.
Prove that $\lim_{n \to \infty} k^nn^p= 0$ where $|k| < 1$ and $p>0$ 
I can see why it should be the case (since exponentials grow faster than polynomials) but I don't really know where to start to try and prove it rigourously. I have identified that it becomes a $"0 \times \infty"$ situation and have tried using L'Hôpital's rule on the expression $k^n/(1/n^p)$ which didn't seem to help. How should go about tackling this problem?


So I tried applying l'Hôpital's rule $p$ times on the expression $k^n/(1/n^p)$
  . I then get $\frac{k^n ln(k)^p}{(-1)^p*p(p+1)(p+2)...(2p)) / n^{2p}}$ and I don't quite see where that gets me.

l'Hôpital is possible, but if you do it that way
$$ \lim_{n\to\infty} k^n n^p =\lim_{n\to\infty}  \frac{k^n}{n^{-p}}=\lim_{n\to\infty} \frac{k^{n}\log k}{(-p)n^{-p-1}} = \dots$$
you end up with terms of the same structure. If instead you use the $\infty/\infty$ l'Hôpital rule, you can reduce the power of $n^p$, and this lets you conclude. Say $p\notin \mathbb N$, then
\begin{align} \lim_{n\to\infty} k^n n^p 
&=\lim_{n\to\infty}  \frac{n^p}{k^{-n}}
\\&= \lim_{n\to\infty} p\frac{n^{p-1}}{k^{-n
} (-\log k)} 
\\&= \lim_{n\to\infty} p(p-1) \frac{n^{p-2}}{k^{-n
} (-\log k)^2} 
\\&\mathrel{ \vdots}\\
& = \lim_{n\to\infty} p(p-1)\dots(p-\lfloor p \rfloor)(-\log k)^{-\lfloor p \rfloor - 1}n^{p-\lfloor p \rfloor - 1} k^n 
\\
&= p(p-1)\dots(p-\lfloor p \rfloor)(-\log k)^{-\lfloor p \rfloor - 1} \lim_{n\to\infty} n^{p-\lfloor p \rfloor - 1} k^n 
\\
&= p(p-1)\dots(p-\lfloor p \rfloor)(-\log k)^{-\lfloor p \rfloor - 1} \lim_{n\to\infty} n^{p-\lfloor p \rfloor - 1}\lim_{n\to\infty} k^n 
\\&=  p(p-1)\dots(p-\lfloor p \rfloor)(-\log k)^{-\lfloor p \rfloor - 1} 0\times 0
\\ &= 0
\end{align}
the proof is easier if $p\in\mathbb N$.
