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$f(x)$ is defined as: $$f(x) = \lim_{n\to\infty}\frac{x^{2n} - 1}{x^{2n} + 1}.$$

Where is $f(x)$ continuous?

Could anyone tell me if my answer is correct and how to improve it?

I use l'Hopital's rule and take successive derivatives. I get $(2n)!/(2n)!$ which is $1$. So, the limit is $1$. I look at special cases of interest $x ∈ (-1,1),x=\pm1,0$.

For $x ∈ (0,1)$, $x^{2n}$ approaches $0$. As well as $-1>x>0$ because $x^{2n}$ is $(x^2)^n$ which cancels out the negative sign. Both gives a answer of $-1$. Not continuous on interval $-1<x<1$.

For $x=0$, $f(0)$ is $-1$. Not continuous at $x=0$.

For $x= \pm1$, the answer is $0$. Not continuous on $x=\pm1$.

For $x>1$ and $x<-1$, they have no surprises.

So, the function is continuous on the intervals $(-\infty, -1)$ and $(1, +\infty)$.

Hagen von Eitzen
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With a bit of thought, you should find (or upon rereading your post you actually found) that $$f(x)=\begin{cases}1&\text{if }{x>1}\\ 0&\text{if }x=1\\ -1&\text{if }-1<x<1\\ 0&\text{if }x=-1\\ 1&\text{if }x<-1\end{cases} $$Thus $f$ is continuous everywhere except at $\pm1$.

Hagen von Eitzen
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  • Why wouldn't it be continuous also at x = (-1,1)? – Math Aug 16 '16 at 15:45
  • @AlanTam Let $a\in (-1,1)$. Let $\epsilon>0$ be given. Pick $\delta = 1-|a|$, so that $\delta>0$. Then for all $x$ with $|x-a|<\delta$ we have $|x|<1$ and hence $f(x)=-1$. Thus $|f(x)-f(a)|=0$, which is of course $<\epsilon$ – Hagen von Eitzen Aug 16 '16 at 15:48
  • Oh, I get it. I was thinking of continuity at a point rather over an interval. Sorry. Thanks. – Math Aug 16 '16 at 15:59