$f(x)=\underset{n→∞}{\lim}\sum\limits_{r=0}^{n-1}\frac{x}{(rx+1)\{(r+1)x+1\}}$, then
Answer & explanation
Correct answer: option 3
$t_{r+1}=\frac{x}{(rx+1)\{(r+1)x+1\}}=\frac{(r+1)x+1-(rx+1)}{(rx+1)[(r+1)x+1]}=\frac{1}{(rx+1)}-\frac{1}{(r+1)x+1}$
$∴S_n=\sum\limits_{r=0}^{n-1}t_{r+1}=\left\{\begin{matrix}1-\frac{1}{nx+1},&x≠0\\0,&x=0\end{matrix}\right.$
$∴f(x)=\underset{n→∞}{\lim}S_n=\left\{\begin{matrix}1,&x≠0\\0,&x=0\end{matrix}\right.$
$∴\underset{x→0}{\lim}f(x)=1$ and f(0) = 0
Hence f (x) is neither continuous nor differentiable at x = 0.
Clearly f (x) is not a periodic function