Proving a function is continuous

**JSB1917** · November 27th 2012, 02:00 PM

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?

**Plato** · November 27th 2012, 02:08 PM

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.

**SworD** · November 27th 2012, 03:34 PM

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.

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