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