JEE Main 2020 — Mathematics Calculus

$$\underset{x\rightarrow 0}{\mathrm{lim}}|\frac{1-x+|x|}{\lambda -x+[x]}|=L$$x→0lim​∣λ−x+[x]1−x+∣x∣​∣=L

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

For left-hand limit

$$x=0-h,x\rightarrow 0,h\rightarrow 0,h>0$$x=0−h,x→0,h→0,h>0

$$LHL=\underset{h\rightarrow 0}{\mathrm{lim}}|\frac{1-(-h)+|-h|}{\lambda -(-h)+[-h]}|$$LHL=h→0lim​∣λ−(−h)+[−h]1−(−h)+∣−h∣​∣

$$\Rightarrow LHL=\underset{h\rightarrow 0}{\mathrm{lim}}|\frac{1+h-(-h)}{\lambda +h+[-h]}|$$⇒LHL=h→0lim​∣λ+h+[−h]1+h−(−h)​∣

Since $h>0\Rightarrow [-h]=-1$h>0⇒[−h]=−1

$$\Rightarrow LHL=\underset{h\rightarrow 0}{\mathrm{lim}}|\frac{1+h+h}{\lambda +h-1}|=\frac{1}{|\lambda -1|}$$⇒LHL=h→0lim​∣λ+h−11+h+h​∣=∣λ−1∣1​

For right-hand limit

$$x=0+h,x\rightarrow 0,h\rightarrow 0,h>0$$x=0+h,x→0,h→0,h>0

$$RHL=\underset{h\rightarrow 0}{\mathrm{lim}}|\frac{1-h+|h|}{\lambda -h+[h]}|$$RHL=h→0lim​∣λ−h+[h]1−h+∣h∣​∣

Since $h>0\Rightarrow [h]=0$h>0⇒[h]=0

$$\Rightarrow RHL=\underset{h\rightarrow 0}{\mathrm{lim}}|\frac{1+h+h}{\lambda +h+0}|=\frac{1}{|\lambda |}$$⇒RHL=h→0lim​∣λ+h+01+h+h​∣=∣λ∣1​

For existence of limit $LHL=RHL$LHL=RHL

$$\Rightarrow \frac{1}{|\lambda -1|}=\frac{1}{|\lambda |}$$⇒∣λ−1∣1​=∣λ∣1​

Cross multiply and square both side.

$$\Rightarrow {\lambda }^{2}={(\lambda -1)}^{2}\Rightarrow {\lambda }^{2}={\lambda }^{2}+1-2\lambda$$⇒λ2=(λ−1)2⇒λ2=λ2+1−2λ

$$\Rightarrow \lambda =\frac{1}{2}$$⇒λ=21​

Limit $L=\frac{1}{\frac{1}{2}}=2$L=21​1​=2