Let x_0 be the real number such that e^{x_0}+x_0=0. For a given real number \alpha, define g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3 (e^x+1) } for all real numbers x. Then which one of the following statements is TRUE? [JEE Adv 2025 P2]

Correct Answer is : For α=3,limx⟶x0|g(x)+ex0x−x0|=0

For \alpha=2, \lim_{x \rightarrow x_0}{| \frac{g(x)+e^{x_0}}{x-x_0} |}=0

For \alpha=2, \lim_{x \rightarrow x_0}{| \frac{g(x)+e^{x_0}}{x-x_0} |}=1

For \alpha=3, \lim_{x \rightarrow x_0}{| \frac{g(x)+e^{x_0}}{x-x_0} |}=0

For \alpha=3, \lim_{x \rightarrow x_0}{| \frac{g(x)+e^{x_0}}{x-x_0} |}=\frac{2}{3}

Correct Answer is : For α=3,limx⟶x0|g(x)+ex0x−x0|=0

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Limits, Continuity and Differentiability

Section: Mathematics

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