Mathematics - Continuity - Differentiability Question with Solution | TestHub

Let $f(x)=\frac{x}{1+x}+\frac{x}{(x+1)(2x+1)}+\frac{x}{(2x+1)(3x+1)}+\dots \infty$ $={\lim }_{n\to \infty }{\sum }_{r=1}^n\frac{x}{[(r-1)x+1](rx+1)}$ $={\lim }_{x\to \infty }{\sum }_{\mathrm{r}=1}^{\mathrm{n}}\left[\frac{\mathrm{x}}{[(\mathrm{r}-1)\mathrm{x}+1]}-\frac{1}{\mathrm{rx}+1}\right]$ $={\lim }_{n\to \infty }\left[1-\frac{1}{nx+1}\right]=1$ For $x=0,$ we have $f(x)=0$ Thus, we have $f(x)=\{\begin{array}{ll}1,& x\ne 0\\ 0,& x=0\end{array}$ Clearly, ${\lim }_{x\to 0^{-}}\mathrm{f}(\mathrm{x})={\lim }_{\mathrm{x}\to 0^{+}}\mathrm{f}(\mathrm{x})\ne \mathrm{f}(0)$ So, $\mathrm{f}(\mathrm{x})$ is not continuous at $\mathrm{x}=0$.