Where is $f$ continuous?
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Where is $f$ continuous?
A function is defined as: \( \displaystyle f(x)=\lim_{n\rightarrow +\infty }\frac{x^{2n}-1}{x^{2n}+1} \). Where is \(f \) continuous?
Hidden Message
Imagination is much more important than knowledge.
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: Where is $f$ continuous?
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,
$$f(x)= \left\{\begin{matrix}
1 &, & x \in (-\infty, -1)\cup (1, +\infty) \\
-1 &, & x \in (-1, 1) \\
0&, & x = \pm 1
\end{matrix}\right.$$
The function \(\displaystyle{f}\) is continuous at \(\displaystyle{\left(-\infty,-1\right)\cup\left(-1,1\right)\cup\left(1,+\infty\right)}\)
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,
$$f(x)= \left\{\begin{matrix}
1 &, & x \in (-\infty, -1)\cup (1, +\infty) \\
-1 &, & x \in (-1, 1) \\
0&, & x = \pm 1
\end{matrix}\right.$$
The function \(\displaystyle{f}\) is continuous at \(\displaystyle{\left(-\infty,-1\right)\cup\left(-1,1\right)\cup\left(1,+\infty\right)}\)
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 27 guests