JEE Main 2020 — Limits Question with Solution

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

Let's find left-hand and right-hand limit.

For left-hand limit

$x=0-h, x\to 0, h\to 0, h>0$

$LHL=\underset{h\to 0}{\lim }\left|\frac{1-\left(-h\right)+\left|-h\right|}{\lambda -\left(-h\right)+\left[-h\right]}\right|$

$\Rightarrow LHL=\underset{h\to 0}{\lim }\left|\frac{1+h-\left(-h\right)}{\lambda +h+\left[-h\right]}\right|$

Since $h>0 \Rightarrow \left[-h\right]=-1$

$\Rightarrow LHL=\underset{h\to 0}{\lim }\left|\frac{1+h+h}{\lambda +h-1}\right|=\frac{1}{\left|\lambda -1\right|}$

For right-hand limit

$x=0+h, x\to 0, h\to 0, h>0$

$RHL=\underset{h\to 0}{\lim }\left|\frac{1-h+\left|h\right|}{\lambda -h+\left[h\right]}\right|$

Since $h>0 \Rightarrow \left[h\right]=0$

$\Rightarrow RHL=\underset{h\to 0}{\lim }\left|\frac{1+h+h}{\lambda +h+0}\right|=\frac{1}{\left|\lambda \right|}$

For existence of limit $LHL=RHL$

$\Rightarrow \frac{1}{\left|\lambda -1\right|}=\frac{1}{\left|\lambda \right|}$

Cross multiply and square both side.

$\Rightarrow {\lambda }^2={\left(\lambda -1\right)}^2 \Rightarrow {\lambda }^2={\lambda }^2+1-2\lambda$

$\Rightarrow \lambda =\frac{1}{2}$

Limit $L=\frac{1}{{\displaystyle \frac{1}{2}}}=2$