Finding the limit of $\lim_{n \to \infty} \frac{\pi^n-n^\pi}{\pi^{n+1}-(n+1)^\pi}$ How can you find the value of $$\lim_{n \to \infty} \frac{\pi^n-n^\pi}{\pi^{n+1}-(n+1)^\pi}$$
I know that the answer is $\dfrac{1}{\pi}$ but I can't figure out how to obtain that. Could someone give me a hint please, thank you
 A: $$ \lim_{n \to \infty} \frac{\pi^n-n^\pi}{\pi^{n+1}-(n+1)^\pi} 
= \lim_{n \to \infty}\frac{1}{\pi} \frac{1-\frac{n^\pi}{\pi^{n}}}{1-\frac{(n+1)^\pi}{\pi^{n+1}}} $$
and observe that $\lim_{n\to\infty}\frac{n^\pi}{\pi^n} = 0$ so the original limit is $\frac{1}{\pi}\frac{1-0}{1-0} = \frac{1}{\pi}$
EDIT: Cuteboy notes that the last limit is not trivial.  The result that for all $A$ and for all $\epsilon>0$, $\frac{x^{A}}{(1+\epsilon)^{x}}\to 0$ as $x\to\infty$ is fundamental to understanding limits of this type.  Exponential growth dominates polynomial growth.  Whether or not you need to add a proof of this to every limit write-up is, I guess, between you and the reader.  If you need a proof, see Cuteboy's answer to this question.
A: The first answers has partially proved your claim. One has still to show that $\lim_{n\to\infty}\frac{n^{\pi}}{\pi^{n}}=0$. It suffices to show $\lim_{n\to\infty}\log\frac{n^{\pi}}{\pi^{n}}=\lim_{n\to\infty}(\pi\log n-n\log\pi)=-\infty$. Now from L'hopital we obtain that $o(1)=\log n/n$, therefore $\pi\log n-n\log\pi=n(o(1)\pi-\log\pi)\to-\infty$ as $n\to\infty$.
