Mathematics - Continuity - Differentiability Question with Solution | TestHub

In the vicinity of $\mathrm{x}=0$, we have ${\mathrm{x}}^2{\sum }_{\mathrm{r}=0}^{\left[\frac{1}{|\mathrm{x}|}\right]}\mathrm{r}={\mathrm{x}}^2\left(1+2+3+\dots \left[\frac{1}{|\mathrm{x}|}\right]\right)$ Use sandwich theorem

$$\begin{array}{rlr}& P=\left(1+2+3+\left[\frac{1}{|\mathrm{x}|}\right]\right)=\frac{{\mathrm{x}}^2\left(1+\left[\frac{1}{|\mathrm{x}|}\right]\right)}{2}\left[\frac{1}{|\mathrm{x}|}\right]& \end{array}$$

$$So\frac{1}{2}(1-|x|)<P\le \frac{1}{2}(1+|x|)$$

Then the limit is $\frac{1}{2}$