Question

Let [t] denote the greatest integer $\le t$. If for some
$\lambda \in R-\{0,1\},\underset{x\to 0}{\lim }\left|\frac{1-x+|x|}{\lambda -x+[x]}\right|=L,$ then $L$ is equal to :

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

Answer 1

Answer: (B)

$LHL:\underset{x\to 0^{-}}{\lim }\left|\frac{1-x-x}{\lambda -x-1}\right|=\left|\frac{1}{\lambda -1}\right|$
$RHL:\underset{x\to 0^{+}}{\lim }\left|\frac{1-x+x}{\lambda -x+1}\right|=\left|\frac{1}{\lambda }\right|$
For existence of limit
$LHL=RHL$
$\Rightarrow \frac{1}{|\lambda -1|}=\frac{1}{|\lambda |}$
$\Rightarrow \lambda =\frac{1}{2}$
$\therefore L=\frac{1}{|\lambda |}=2$