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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If f(x)=x⋅e∣x∣+[x]−2∣x∣+[x]f(x)=x\cdot\dfrac{e^{|x|+[x]}-2}{|x|+[x]} where [⋅][\cdot] is GIF, then
Alim⁡x→0+f(x)=−1\displaystyle\lim_{x\to 0^{+}}f(x)=-1correct
Blim⁡x→0−f(x)=0\displaystyle\lim_{x\to 0^{-}}f(x)=0correct
Clim⁡x→0f(x)=−1\displaystyle\lim_{x\to 0}f(x)=-1
Dlim⁡x→0f(x)=0\displaystyle\lim_{x\to 0}f(x)=0
Solution
Step 1: Compute the RHL (x→0+x\to 0^{+}). For small x>0x>0: ∣x∣=x|x|=x and [x]=0[x]=0, so ∣x∣+[x]=x|x|+[x]=x:
lim⁡x→0+x⋅ex−2x=lim⁡x→0+(ex−2)=1−2=−1\lim_{x\to 0^{+}}x\cdot\dfrac{e^{x}-2}{x}=\lim_{x\to 0^{+}}(e^{x}-2)=1-2=-1
 So RHL =−1=-1. Option (1) is correct. Step 2: Compute the LHL (x→0−x\to 0^{-}). For small x<0x<0: ∣x∣=−x|x|=-x and [x]=−1[x]=-1, so ∣x∣+[x]=−x−1|x|+[x]=-x-1:
lim⁡x→0−x⋅e−x−1−2−x−1\lim_{x\to 0^{-}}x\cdot\dfrac{e^{-x-1}-2}{-x-1}
 As x→0−x\to 0^{-}: e−x−1→e−1e^{-x-1}\to e^{-1}, −x−1→−1-x-1\to -1, so the fraction →e−1−2−1=2−e−1\to\dfrac{e^{-1}-2}{-1}=2-e^{-1}. Multiplying by x→0x\to 0:
LHL=0⋅(2−e−1)=0\text{LHL}=0\cdot(2-e^{-1})=0
 So LHL =0=0. Option (2) is correct. Step 3: LHL ≠\neq RHL, so the two-sided limit does not exist. Options (3) and (4) are wrong. Correct answers: (1) and (2)
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