Question

Let $f\left(x\right)=x\left[\frac{1}{x}\right]$ for all $x(\ne 0)\in R$, where for each $t\in R$, $\left[t\right]$ denotes the greatest integer less than or equal to t. Then?

• A. $\underset{x\to 0+}{lim}f\left(x\right)=0$
• B. $\underset{x\to \frac{1}{2}+}{lim}f\left(x\right)=1$
• C. $\underset{x\to \frac{1}{2}-}{lim}f\left(x\right)=1$
• D. $\underset{x\to 2-}{lim}f\left(x\right)=1$

Answer 1

The correct option is A $\underset{x\to 0+}{lim}f\left(x\right)=0$

$f\left(x\right)=x\left[\frac{1}{x}\right]$

$f\left(x\right)x\left[\frac{1}{x}\right]$

for $[x{]}_{mx\to 0}$ Let $x=0.1$

$\therefore \left[x\right]$ will return $o$

$\therefore \left[\frac{1}{x}\right]$ if $x=0.1$

$\left[\frac{1}{x}\right]=10$

$\therefore f(x{)}_{mx\to 0}x\left[\frac{1}{x}\right]=0\times \left[\frac{1}{x}\right]$

$=0$