# Evaluating a Simple Limit as $n$ Approaches Infinity

I apologize in advance for the simplicity of this question, I'm sure I'm missing something right in front of me.

For $$z < r$$, how can we show that the following limit evaluates to zero?

We have $$\alpha = |z/r| <1$$, so $$\alpha = 1/(1+\beta)$$ where $$\beta > 0$$.
For all $$n >2$$, by the binomial theorem,
$$0 \leqslant n\alpha^{n-1} = \frac{n}{(1+ \beta)^{n-1}} < \frac{n}{\frac{1}{2}(n-1)(n-2)\beta^2} \rightarrow_{n \to \infty} 0$$
If $$a<0$$,$$\lim_{x\to\infty}xe^{a(x-1)}=\lim_{x\to\infty}\frac x{e^{-a(x-1)}}=0,$$by L'Hopital's rule. Now, apply this with $$a=\log\left\lvert\frac zr\right\rvert<0$$.
• Since $a<0$, it is actually $\frac\infty\infty$. And then you apply L'Hopital's Rule. – José Carlos Santos Aug 15 at 14:09