Continuous function

[Tolaso J Kos]

A function is defined as: \( \displaystyle f(x)=\lim_{n\rightarrow \infty }\frac{x^{2n}-1}{x^{2n}+1} \). Where is \(f \) continuous?



P.S I don't have a solution to this.

[Papapetros Vaggelis]

For each \(\displaystyle{x\in\mathbb{R}}\) holds : \(\displaystyle{\dfrac{x^{2\,n}-1}{x^{2\,n}+1}=\dfrac{\left(x^2\right)^{n}-1}{\left(x^2\right)^{n}+1}}\) .

Let \(\displaystyle{x\in\mathbb{R}}\) . If \(\displaystyle{x\in\left(-\infty,-1\right)\cup\left(1,+\infty\right)}\) , then \(\displaystyle{x^2>1}\) and

\(\displaystyle{\lim_{n\to \infty}\left(x^2\right)^{n}=\infty}\), so

\(\displaystyle{f(x)=\lim_{n\to \infty}\dfrac{\displaystyle{1-\dfrac{1}{x^{2\,n}}}}{\displaystyle{1+\dfrac{1}{x^{2\,n}}}}=1}\) .

If \(\displaystyle{x\in\left(-1,1\right)}\) , then \(\displaystyle{0\leq x^2<1}\) and \(\displaystyle{\lim_{n\to \infty}\left(x^2\right)^{n}=0}\) .

Therefore, \(\displaystyle{f(x)=\lim_{n\to \infty}\dfrac{\left(x^2\right)^{n}-1}{\left(x^2\right)^{n}+1}=-1}\) .

Finally, if \(\displaystyle{x\in\left\{-1,1\right\}}\) , then \(\displaystyle{x^{2\,n}-1=0}\) and \(\displaystyle{f(x)=0}\) .

So,

\(\displaystyle{f(x)=\begin{cases}
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x\in\left(-\infty,-1\right)\cup\left(1,+\infty\right)\\
-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x\in\left(-1,1\right)\\
0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x=-1,1
\end{cases}}\)

The function \(\displaystyle{f}\) is continuous at \(\displaystyle{\left(-\infty,-1\right)\cup\left(-1,1\right)\cup\left(1,+\infty\right)}\)