# Proving a function is continuous

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

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

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

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 $\displaystyle f(x)=kx$ is continuous for any real number $\displaystyle k$.

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

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

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

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

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

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