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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.

emka
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    Is there a question? – wj32 Dec 04 '12 at 05:09
  • Here is a related problem. – Mhenni Benghorbal Dec 04 '12 at 05:27
  • Does it matter that the function goes from $f: \mathbb{R}^n \to \mathbb{R}$? – emka Dec 04 '12 at 05:29
  • 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 04 '12 at 05:33
  • @MhenniBenghorbal Is my answer completely wrong? I'm trying to make sure I understand how to apply the definition of continuity properly. – emka Dec 04 '12 at 05:56
  • This question is a good example where continuity in higher dimension is exactly as easy to handle as in dimension 1 (for every norm). If you replace $|x|$, $|x_0|$ and $|x-x_0|$ by $|x|$, $|x_0|$ and $|x-x_0|$, and if you use the triangular inequality for the norm $|\ |$ like you did for $|\ |$, then your proof will be perfectly valid. – Did Dec 04 '12 at 06:33
  • Thanks. I figured he did the $\mathbb{R}^n$ to toy with your minds a bit. – emka Dec 04 '12 at 06:57
  • Your current answer is absolutely correct, but since you are not sure about it, you should tell us at which step you have doubts. – Phira Dec 04 '12 at 16:01

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