Question

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

• A. Has no limit
• B. Is not continuous
• C. Is continuous but not differentiable
• D. Is differentiable

Answer 1

The correct option is B Is not continuous

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