Question

If $f\left(x\right)=\frac{\left[x\right]}{\left|x\right|},x\ne 0$ where [.] denotes the greatest integer function, then $f^{{}^{\prime }}\left(1\right)$ is

• A. $-1$
• B. $\infty$
• C. non existent
• D. none of these

Answer 1

The correct option is C non existent

$f^{\prime }(1+0)=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}=\underset{h\to 0}{lim}\frac{\frac{[1+h]}{|1+h|}-\frac{\left[1\right]}{\left|1\right|}}{h}=\underset{h\to 0}{lim}\frac{\frac{1}{1+h}-1}{h}=\underset{h\to 0}{lim}\frac{-h}{h(1+h)}=-1f^{\prime }(1-0)=\underset{h\to 0}{lim}\frac{f(1-h)-f\left(1\right)}{-h}=\underset{h\to 0}{lim}\frac{\frac{[1-h]}{|1-h|}-\frac{\left[1\right]}{\left|1\right|}}{-h}=\underset{h\to 0}{lim}\frac{0-1}{-h}=\infty$