Question

If $f\left(x\right)=\frac{|x|}{\left[x\right]}$ and $x\notin [0,1]$ , where $\left[x\right]$ denotes greatest integer function, then$\underset{x\to 2^{+}}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}?$

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$\frac{1}{2}$

As $x\to 2^{+}$

$2<x<3$

Or

$\left[x\right]=2$

Thus, $f\left(x\right)=\frac{|x|}{2}$

$f\left(2\right)=1$

Hence,

$\underset{x\to 2^{+}}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}$

Or,$\underset{x\to 2^{+}}{lim}\frac{\frac{|x|}{2}-1}{x-2}$

Since, it is of $\frac{0}{0}$ form

We apply L-Hospital Rule,

So,$\underset{x\to 2^{+}}{lim}{f}^{\prime }\left(x\right)=\frac{1}{2}$