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Limits: Exists Exists Equal

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

Question
If lim⁡x→0g(x)\displaystyle\lim_{x\to 0}g(x) exists and lim⁡x→0g(x)⋅(e1/x−e−1/xe1/x+e−1/x)\displaystyle\lim_{x\to 0}g(x)\cdot\left(\dfrac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right) also exists, then g(x)g(x) is(are) equal to
Axxcorrect
Bx+4x+4
Cx2x^{2}correct
D4x2+3x+24x^{2}+3x+2
Solution
Step 1: Let f(x)=e1/x−e−1/xe1/x+e−1/xf(x)=\dfrac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}. Step 2: Compute the LHL. As x→0−x\to 0^{-}, 1x→−∞\dfrac{1}{x}\to -\infty, so e1/x→0e^{1/x}\to 0 while e−1/x→∞e^{-1/x}\to\infty. Multiply numerator and denominator by e1/xe^{1/x}:
f(x)=e2/x−1e2/x+1  →  0−10+1=−1f(x)=\dfrac{e^{2/x}-1}{e^{2/x}+1}\;\to\;\dfrac{0-1}{0+1}=-1
 Step 3: Compute the RHL. As x→0+x\to 0^{+}, 1x→+∞\dfrac{1}{x}\to +\infty, so e−1/x→0e^{-1/x}\to 0 while e1/x→∞e^{1/x}\to\infty. Multiply numerator and denominator by e−1/xe^{-1/x}:
f(x)=1−e−2/x1+e−2/x  →  1−01+0=1f(x)=\dfrac{1-e^{-2/x}}{1+e^{-2/x}}\;\to\;\dfrac{1-0}{1+0}=1
 Step 4: So LHL =−1=-1 and RHL =1=1, and f(x)f(x) has a jump discontinuity at x=0x=0. Step 5: For lim⁡x→0g(x)⋅f(x)\displaystyle\lim_{x\to 0}g(x)\cdot f(x) to exist, we need
lim⁡x→0−g(x)⋅(−1)=lim⁡x→0+g(x)⋅1\lim_{x\to 0^{-}}g(x)\cdot(-1)=\lim_{x\to 0^{+}}g(x)\cdot 1
 Since lim⁡x→0g(x)\displaystyle\lim_{x\to 0}g(x) exists (call it LL), this becomes −L=L  ⇒  L=0-L=L\;\Rightarrow\;L=0. Step 6: So we need g(0)=0g(0)=0 (i.e., gg vanishes at 00). Check options: (A) g(x)=xg(x)=x: g(0)=0g(0)=0. Valid. (B) g(x)=x+4g(x)=x+4: g(0)=4≠0g(0)=4\neq 0. Invalid. (C) g(x)=x2g(x)=x^{2}: g(0)=0g(0)=0. Valid. (D) g(x)=4x2+3x+2g(x)=4x^{2}+3x+2: g(0)=2≠0g(0)=2\neq 0. Invalid. Correct answers: (1) and (3)
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