# Showing a function is continuous from the definition

I am trying to recall this from memory from an exam that I have lost. I'm wondering if this approach is correct - I most likely fumbled this up during the exam.

Let $f: \mathbb{R}^n \to \mathbb{R}$ be defined as $f(x)=3|x|+5$. Show this is continuous.

Let $\epsilon > 0$ Fix $x_0$. Then for $|x-x_0|< \delta$.

$|f(x)-f(x_0)|=|3|x|-5-3|x_0|+5|=3||x|-|x_0|| \leq 3|x-x_0|<3\delta$

If $\delta = \frac{\epsilon}{3}$ we get what we desired.

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Is there a question? –  wj32 Dec 4 '12 at 5:09
Thank you. Now there is. –  ortl Dec 4 '12 at 5:13
Here is a related problem. –  Mhenni Benghorbal Dec 4 '12 at 5:27
Does it matter that the function goes from $f: \mathbb{R}^n \to \mathbb{R}$? –  ortl Dec 4 '12 at 5:29
First, see functions with two variables and see what happens. Continuity in higher dimensions is harder to handle than in one dimension. –  Mhenni Benghorbal Dec 4 '12 at 5:33