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?

Re: Proving a function is continuous

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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.

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.