Q46 / 125 · BITSAT 2015 · Mathematics

If a function $f(x)$ is given by $f(x)=\frac{x}{1+x}+\frac{x}{(x+1)(2 x+1)}$ $+\frac{x}{(2 x+1)(3 x+1)}+\ldots, \infty$, then at $x=0, f(x)$

  • A
    has no limit
  • Correct answer
    B
    is not continuous
  • C
    is continuous but not differentiable
  • D
    is differentiable

Solution

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

Hence, the correct answer is option: 'B'.

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