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Find $$\lim_{x\rightarrow 0^{+}}x\cdot \bigg(\bigg\lfloor \frac{1}{x}\bigg\rfloor+\bigg\lfloor \frac{2}{x}\bigg\rfloor+\bigg\lfloor \frac{3}{x}\bigg\rfloor+\cdots \cdots +\bigg\lfloor \frac{15}{x}\bigg\rfloor\bigg)$$ where $\lfloor x\rfloor$ is an integer part of $x$.

I have solved it using squeeze theorem. Could some help me how to solve it without squeeze theorem thanks

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  • $\begingroup$ write $[k]$ as $k -(x) $ and continue. where () is fractional part $\endgroup$ – Akul Singhal Apr 9 '18 at 14:48
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Observe that $\lfloor y\rfloor = y+\theta(y)$ where $\theta(y)$ is bounded. Hence $$ x\cdot\left \lfloor\frac nx\right \rfloor = n+x\theta(n/x) $$ As $\theta $ is bounded, we have $x\theta(n/x)\to 0$ as $x\to 0$, whence the result - but then again, this last step is still squeezing in disguise.

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This is one method you can do

$ x\times[\frac{r}{x}] = x\times(\frac{r}{x} - $ $\{\frac{r}{x}\})$ where {} is fractional part function.

Now $x\times\frac{r}{x} = x$ and $x\times\{\frac{r}{x}\} \rightarrow 0$ as $ x \rightarrow 0 $ and $\{\frac{r}{x}\}$ is a finite number between 0 and 1.

So your limit becomes

$ \sum\limits_{r=1}^{15} r$

Which I think you can easily calculate and is equal to 120.

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Let $\displaystyle \frac{k}{x}=n+r$, where $\displaystyle n=\left\lfloor \frac{k}{x}\right\rfloor$. Then $\displaystyle x\left\lfloor \frac{k}{x}\right\rfloor=\frac{kn}{n+r}$

$$k\ge x\left\lfloor \frac{k}{x}\right\rfloor>\frac{kn}{n+1}=k-\frac{k}{n+1}$$

As $x\to 0^+$, $n\to\infty$ and hence $\displaystyle x\left\lfloor \frac{k}{x}\right\rfloor\to k$.

So, $\displaystyle \lim_{x\to0^+}x\sum_{k=1}^{15}\left\lfloor \frac{k}{x}\right\rfloor=\sum_{k=1}^{15}k=120$.

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If $0\lt x\lt\delta=\epsilon/15$, then, using $120=1+2+\cdots+15$ and $|\lfloor u\rfloor-u|\le1$ for all $u$, we have

$$\begin{align} \left|x\left(\left\lfloor1\over x \right\rfloor+\left\lfloor2\over x \right\rfloor+\cdots+\left\lfloor15\over x \right\rfloor \right)-120\right| &=\left|x\left(\left\lfloor1\over x \right\rfloor-{1\over x}\right)+x \left(\left\lfloor2\over x \right\rfloor-{2\over x}\right)+\cdots+x\left(\left\lfloor15\over x \right\rfloor-{15\over x} \right)\right|\\ &\le|x|+|x|+\cdots+|x|=15x\lt15\delta=\epsilon \end{align}$$

and thus, by definition,

$$\lim_{x\to0^+}x\left(\left\lfloor1\over x \right\rfloor+\left\lfloor2\over x \right\rfloor+\cdots+\left\lfloor15\over x \right\rfloor \right)=120$$

Remark: The restriction to $x\to0^+$ is irrelevant; the proof works for $x\to0$ from either side.

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