Practicing Success
If $f(x)=\frac{[x]}{|x|}, x \neq 0$ where [.] denotes the greatest integer function, then f'(1) is |
-1 1 non-existent $\infty$ |
non-existent |
We have, $f(x)=\frac{[x]}{|x|}= \begin{cases}0, & 0<x<1 \\ \frac{1}{x}, & 1 \leq x<2\end{cases}$ Clearly, $\lim\limits_{x \rightarrow 1^{-}} f(x)=0$ and $\lim\limits_{x \rightarrow 1^{+}} f(x)=1$ ∴ $\lim\limits_{x \rightarrow 1^{-}} f(x) \neq \lim\limits_{x \rightarrow 1^{+}} f(x)$ Thus, f(x) is not continuous at x = 1. Hence, f(x) is not differentiable at x = 1. Consequently, f'(1) does not exist. |