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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.

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    Is there a question?2012-12-04
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    Here is a [related problem](http://math.stackexchange.com/questions/209440/how-to-show-that-fx-x2-is-continuous-at-x-1/209492#209492).2012-12-04
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    Does it matter that the function goes from $f: \mathbb{R}^n \to \mathbb{R}$?2012-12-04
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    First, see functions with two variables and see what happens. Continuity in higher dimensions is harder to handle than in one dimension.2012-12-04
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    @MhenniBenghorbal Is my answer completely wrong? I'm trying to make sure I understand how to apply the definition of continuity properly.2012-12-04
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    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.2012-12-04
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    Thanks. I figured he did the $\mathbb{R}^n$ to toy with your minds a bit.2012-12-04
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    Your current answer is absolutely correct, but since you are not sure about it, you should tell us at which step you have doubts.2012-12-04

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