Given below are two statements:
Statement I:
$lim x \to 0 (\frac{tan ^{- 1} x + log _{e} \frac{1 + x}{1 - x} - 2 x}{x ^{5}}) = \frac{2}{5}$
Statement II: $lim x \to 1 (x^{\frac{2}{1 - x}}) = \frac{1}{e ^{2}}$
In the light of the above statements, choose the correct answer from the options given below:

Question

Given below are two statements:
Statement I:

lim x \to 0 (\frac{tan ^{- 1} x + log _{e} \frac{1 + x}{1 - x} - 2 x}{x ^{5}}) = \frac{2}{5}

Statement II:

lim x \to 1 (x^{\frac{2}{1 - x}}) = \frac{1}{e ^{2}}

In the light of the above statements, choose the correct answer from the options given below:

Accepted Answer

Correct answer: C. (c) limx 0 0.05em tan^-1 ⁡x+ 12 [ ln ⁡ (1+x )- ln ⁡ (1-x ) ]-2xx^5 = limx 0 0.05em (x- x^33+ x^55 ⋯)+ 12(x- x^22+ x^33 … .- ((-x- x^22- x^33 ⋯) )-2x)x^5 = limx 0 0.05em 2x+ 2x^55 ⋯-2xx^5= 25 limx 1 0.05emx^ 2 (1-x )=e^ limx 1 0.05em ( 21-x ) (x-1 )=e^-2 ⇒ Both statements correct

$Given below are two statements: Statement I: lim x \to 0 (\frac{tan ^{- 1} x + log _{e} \frac{1 + x}{1 - x} - 2 x}{x ^{5}}) = \frac{2}{5} Statement II: lim x \to 1 (x^{\frac{2}{1 - x}}) = \frac{1}{e ^{2}} In the light of the above statements, choose the correct answer from the options given below:$

Solution

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