Question:
Let $[t]$ denote the greatest integer $\leq t$. If for some
$\lambda \in \mathbf{R}-\{0,1\}, \lim _{x \rightarrow 0}\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|=L$, then $L$ is equal to $:$
Correct Option: , 2
Solution:
Given $\lim _{x \rightarrow 0}\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|=L$
Here, L.H.L. $=\lim _{h \rightarrow 0}\left|\frac{1+h+h}{\lambda+h-1}\right|=\left|\frac{1}{\lambda-1}\right|$
R.H.L. $=\lim _{h \rightarrow 0}\left|\frac{1-h+h}{\lambda+h+0}\right|=\left|\frac{1}{\lambda}\right|$
Given that limit exists. Hence L.H.L. = R.H.L.
$\Rightarrow|\lambda-1|=|\lambda|$
$\Rightarrow \lambda=\frac{1}{2}$ and $L=\left|\frac{1}{\lambda}\right|=2$