If $f(x)=\frac{[x]}{|x|}, x \neq 0$, where [.] denotes the greatest integer function, then f'(1) is

non-existent

We have,

$f(x)=\frac{[x]}{|x|}= \begin{cases}\frac{-1}{|x|}=\frac{1}{x}, & \text { if }-1<x<0 \\ \frac{0}{|x|}=0, & \text { if } 0<x<1 \\ \frac{1}{|x|}=\frac{1}{x}, & \text { if } 1 \leq x<2\end{cases}$

∴ (LHD at x = 1) = $\lim\limits_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim\limits_{x \rightarrow 1^{-}} \frac{0-1}{x-1} \rightarrow+\infty$

and, (RHD at x = 1) = $\lim\limits_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim\limits_{x \rightarrow 1^{+}} \frac{\frac{1}{x}-1}{x-1}$

$=\lim\limits_{x \rightarrow 1} \frac{1-x}{x(x-1)}=-1$

Clearly, (LHD at x = 1) ≠ (RHD at x = 1)

So, f'(1) does not exist.