# Proving a function is continuous

• Nov 27th 2012, 02:00 PM
JSB1917
Proving a function is continuous
Problem: Prove using epilson delta definition that $f(x)=kx$ is continuous for any real number $k$.

I have the $k \neq 0$ case. I just had some problems with the $k=0$ case. Here's what I got.

Let $\epsilon > 0$ and choose $\delta=\epsilon$, then $|x-x_{0}| < \delta$ implies that $|x-x_{0}| < \epsilon$ and so $|k||x-x_{0}| < |k|\epsilon$. Then $|kx-kx_{0}| < |k|\epsilon$ and since $k=0$ and $f(x)=kx=0$ and $f(x_{0})=kx_{0}=0$ then $|f(x)-f(x_{0})| < |k|\epsilon$. Since $\epsilon > 0$, then $\epsilon > |k|\epsilon$. So $|f(x)-f(x_{0})| < \epsilon$.

Thoughts? Corrections?
• Nov 27th 2012, 02:08 PM
Plato
Re: Proving a function is continuous
Quote:

Originally Posted by JSB1917
Problem: Prove using epilson delta definition that $f(x)=kx$ is continuous for any real number $k$.

I have the $k \neq 0$ case. I just had some problems with the $k=0$ case.

Oh, come on. If $k=0$ then $f(x)=0$ everywhere.

Thus $|x-a|<\delta$ implies $|0-0|<\epsilon$ everywhere.
• Nov 27th 2012, 03:34 PM
SworD
Re: Proving a function is continuous
Quote:

Originally Posted by Plato
Oh, come on. If $k=0$ then $f(x)=0$ everywhere.

Thus $|x-a|<\delta$ implies $|0-0|<\epsilon$ everywhere.

Lol, this is the problem with drilling too much rigor into people. They start losing their sense of intuition.