Is the function continuous at x=0?

Question

Check if the function $f$ is continuous. $f(x)=$\begin{matrix} 0 & ,x=0\\ \frac{1}{[\frac{1}{x}]} & ,0<x\leq 1 \end{matrix}.

For $0<x\leq 1$,,f is continuous because it is fraction of continuous functions. How can I check if it is continuous at $x=0$?

good to see that $x\in (0,1]\Longleftrightarrow \frac{1}{x}\in [1,+\infty)$ — Mikasa, Feb 09 '14 at 16:07

Ishfaaq · Answer 1 · 2014-02-09T17:19:06.883

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$$|f(x) - f(0)| = |\frac 1 {[\frac 1 x]} - 0| = |\frac 1 {[\frac 1 x]}| = \frac 1 {[\frac 1 x]} \le \frac 1 {\frac 1 x} = x = x - 0\;\; (\text {since $x > 0$ and $\frac 1 {[\frac 1 x]} \le \frac 1 {\frac 1 x}$})$$

Given any $\epsilon \gt 0 $ however small, $\exists \ \delta( = \epsilon) \gt 0 $ such that $|f(x) - f(0)| \lt \epsilon$ whenever $x = x - 0 \lt \delta$

$\implies \lim_{x \rightarrow 0^+} f(x) = f(0)\implies f$ is right-continuous at $x = 0$.

edited Feb 09 '14 at 17:19

answered Feb 09 '14 at 16:43

Ishfaaq

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In the denominator we have [1/x] in the function rules. :) – Mikasa Feb 09 '14 at 16:44
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@B. S. : Not following that.. Did I miss something? Could you explain please.. – Ishfaaq Feb 09 '14 at 16:47
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[1/x] is the integer part of 1/x.. – evinda Feb 09 '14 at 16:49
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Alright.. Then isn't it true that $\frac 1 {[\frac 1 x]} \le \frac 1 {\frac 1 x}$ – Ishfaaq Feb 09 '14 at 16:52
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@Ishfaaq Yes!And do we also know that $\frac{1}{[\frac{1}{x}]}\geq 0$ ? – evinda Feb 09 '14 at 17:09
@evinda: Actually $ 0 \lt x \le 1 \implies 1 \le \frac 1 x \implies 1 \le [\frac{1}{x}] \implies \frac{1}{[\frac{1}{x}]} \ge 0$ – Ishfaaq Feb 09 '14 at 17:12
And @evinda that made me realise my wording wasn't too great. I edited my answer again. Looks alright now me thinks.. – Ishfaaq Feb 09 '14 at 17:20

Is the function continuous at x=0?

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