# Continuous function #1

• Jul 9th 2010, 09:53 PM
Also sprach Zarathustra
Continuous function #1
Prove:

If $\displaystyle f(x)$ is continius function on $\displaystyle [0,\infty)$ and if $\displaystyle lim_{x\to \infty} f(x)=M<\infty$, then $\displaystyle f(x)$ is bounded on$\displaystyle [0,\infty]$

What I did..

I looked the next two intervals:

1. $\displaystyle [0,M]$, $\displaystyle f(x)$ is bounded there.

2. $\displaystyle [M, \infty)$, and I don't know what to do next here...

Thanks!
• Jul 10th 2010, 12:04 AM
Opalg
Quote:

Originally Posted by Also sprach Zarathustra
Prove:

If $\displaystyle f(x)$ is continius function on $\displaystyle [0,\infty)$ and if $\displaystyle lim_{x\to \infty} f(x)=M<\infty$, then $\displaystyle f(x)$ is bounded on$\displaystyle [0,\infty]$

What I did..

I looked the next two intervals:

1. $\displaystyle [0,M]$, $\displaystyle f(x)$ is bounded there.

2. $\displaystyle [M, \infty)$, and I don't know what to do next here...

Thanks!

Use the definition of $\displaystyle \lim_{x\to \infty} f(x)=M$, taking $\displaystyle \varepsilon=1$. It tells you that there exists N such that $\displaystyle |f(x)-M|<1$ whenever x>N. So f is bounded on $\displaystyle [N,\infty)$. But it is also bounded on [0,N] ... .