Deriving limits when n goes to infinity How do i evaluate the limit below?
$$\lim_{n\to\infty}\frac{1}{1-x^{n+1}}=\left\{\begin{array}\\
&\text{if}&|x|<1\\
&\text{if}&|x|>1
\end{array}\right.$$
 A: The answer is:  1, if /x/ < 1 and 0 if /x/ > 1. And this follows from the fact that:
/x/^n --> 0 when /x/ < 1 and --> infinity if /x/ > 1 as n --> infinity.
A: $$\lim_{n\to\infty}\frac{1}{1-x^{n+1}}$$
Let's do the first part, if $|x|<1$. Let's say it's $0.5$ (positive or negative won't matter at the infinity scale):
$$\lim_{n\to\infty}\frac{1}{1-0.5^{n+1}}$$
Focus on $0.5^{n+1}$:
$$0.5^{n+1}=\frac{1}{2^{n+1}}$$
And of course, as $n$ approaches $\infty$:
$$\frac{1}{2^{\infty+1}}=0$$
{The denominator will get so large, it will just be 0.}
Now plug this result back in!
$$\lim_{n\to\infty}\frac{1}{1-0.5^{n+1}}=\frac{1}{1-0}=1$$
Alright, now time to do it for $|x| > 1$. Take $x=5$ for example:
$$\lim_{n\to\infty}\frac{1}{1-5^{n+1}}$$
Look at the bottom again. We know that:
$$\lim_{n\to\infty} \pm5^{n+1}=\pm\infty$$
{Positive or negative won't really matter here either.}
Plugging it in back, we see:
$$\lim_{n\to\infty}\frac{1}{1-5^{n+1}}=\frac{1}{1\mp\infty}$$
Again, always look at the denominator. $1\mp\infty$ will be either positive or negative $\infty$. Since there is $\infty$ the denominator, this will yield to $0$.
If you have any questions or need further explanations, comment below~.
Cheers!
-Shahar
