Question

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

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

Answer 1

The correct option is B $2$

Given:
$\underset{x\to 0}{lim}\left|\frac{1-x+\left|x\right|}{\lambda -x+\left[x\right]}\right|=L$

R.H.L.:
$\underset{x\to 0^{+}}{lim}\left|\frac{1-x+\left|x\right|}{\lambda -x+\left[x\right]}\right|=\left|\frac{1-x+x}{\lambda -x+0}\right|=\frac{1}{|\lambda |}$

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