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Is the function $g:\mathbb{R} \setminus\{0\} \to \mathbb{R}$ given by $g(x) = 1/x^3$ continuous? Why or why not?

A real valued function $f$ is continuous at $a \in \mathbb R$ if the $\lim_{x \to a}f(x) = f(a)$

And more formally a function is continuous at x=a if there is a number $\epsilon>0$ and $\delta>0$ such that there is an interval I: $f(a) \pm \epsilon$ such that I maps the interval M: $a \pm \delta$.

My problem is i only know how to do the first proof and it's about continuity to a point!

But of course by inspection the domain of $1/x^3$ $\{x\in \mathbb {R} |\, x \neq 0 \}$. So with parameters excluding x=0 f(x) is continuous for all values specified in the domain.

Ori
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Ivan
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    Prove that $x \mapsto x^3$ and $x \mapsto 1/x$ are continuous. This is a simpler task. –  Aug 18 '15 at 01:36
  • "And more formally a function is continuous at $x=a$ if there is a number $\epsilon>0$ and $\delta>0$..." This is not correct. It should be "For each $\epsilon>0$ there exists a $\delta_\epsilon>0$ such that $f$ maps the interval $a\pm\delta_\epsilon$ into the interval $f(a)\pm\epsilon$" For the second part, pick a general point $a\in\mathbb{R}\setminus{0}$ and then work from there. If you can show that a function is continuous at every point in its domain, you have shown that it is continuous. – Ori Aug 18 '15 at 01:44

2 Answers2

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Let $\varepsilon, c > 0$. Since $$ |g(x) - g(c)| = \frac{|x-c||x^{2}+xc+c^{2}|}{x^{3}c^{3}} < \frac{|x-c|38}{c^{4}} < \varepsilon $$ if $|x-c| < c/2$ and $< c^{4}\varepsilon/38$, taking $\delta := \min \{c/2, c^{4}\varepsilon/38 \}$ suffices.

By symmetry the function $x \mapsto 1/x^{3}$ is continuous on $]-\infty, 0[$.

Yes
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I'm not clear what your question is! You apparently are saying that you know how to prove that a function is continuous at a point but not how to prove a function "is continuous" on it domain? Is that it? The definition of "continuous on a set" is "continuous at every point of that set".

So take some general $x_0$ and prove the function is continuous at that $x_0$. As long as you do not assume any particular properties for $x_0$- except, of course, that it is not $0$- that will prove it is continuous for all x except $x=0$.

Michael Dyrud
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user247327
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  • Hey Michael, It's the one on the title and highlighted. "Is the function g:ℝ∖{0}→ℝ given by g(x)=1/x3 continuous? Why or why not?" the other part is just my notes – Ivan Aug 18 '15 at 03:25
  • So prove that as x approaches a f(x) equal a? For some point. Is that it? – Ivan Aug 18 '15 at 23:51