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Prove that if $\lim\limits_{x\to 0}f\bigg(x\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg)\bigg)$, then $\lim\limits_{x\to 0}f(x)=0$

My attempt:

If I can show that $\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg) \to 1$ as $x\to 0$, then we are done.

We know, $n\le \dfrac{1}{x}\le n+1$, so $\bigg\lfloor\dfrac{1}{x}\bigg\rfloor=n$.

But, I cannot do anything more to it. Please help. Thank you.

dustin
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Swadhin
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2 Answers2

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Use the Squeeze Theorem. First note that

$$\frac{1}{x}-1<\left\lfloor\frac{1}{x}\right\rfloor\le\frac{1}{x}$$

from which we obtain

$$ \left|x\right|\ge \left|x\right|\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\ge 0$$

Therefore, as $x\to 0$, the Squeeze Theorem guarantees that

$$\lim_{x\to 0} x\,\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)=0 $$

which implies

$$\lim_{x\to 0}f\left(\,x\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right)=\lim_{x\to 0}f(x)$$

If $f$ is continuous, then $$\lim_{x \to 0} f\left(x \left(\frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right)=f\left(\lim_{x \to 0} \left[x \left(\frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right]\right)=f(0). $$

Mark Viola
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  • This might help you out buddy. –  Mar 04 '15 at 06:19
  • I just made an edit to it. Note how the delimiters are scaled and that you can actually use $\lim$ to produce $\lim$ instead of the funny looking $lim$ where you get $lim$. –  Mar 04 '15 at 06:36
  • Cool! Pretty fun isn't it? That tutorial I shared with you is quite nice, for it really shows a lot of what you can do. Also, if you hover your cursor over some mathematics and right-click...you can do Show Math As > TeX Commands ... this will show you how someone typeset what they did. –  Mar 04 '15 at 06:43
  • For example, you can check out this question to see how the table was typeset and how everything else is typeset. The main thing running it all is $\rm\LaTeX$, but MathJax is what's ultimately used on MSE, and sometimes conventional $\rm\LaTeX$ is rendered properly and sometimes it isn't. But that tutorial shows you most of what you need to know. –  Mar 04 '15 at 06:45
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$$\lim_{x\to 0} {x\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg)}=\lim_{x\to 0} 1-x\bigg\lfloor\dfrac{1}{x}\bigg\rfloor$$ But $$\lim_{x\to 0}x\bigg\lfloor\dfrac{1}{x}\bigg\rfloor =1$$(Use the sequeeze theorem and the fact that $$\frac{1}{x}-1<[\frac{1}{x}]\le\frac{1}{x}$$ Therefore $$\lim_{x\to 0} f\bigg({x\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg)\bigg)}=\lim_{t\to 0}f(t)$$ in the latter, if the original limit is zero, so is the limit in RHS.

Fermat
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