Given below are two statements: Statement I : lim x->0 (tan^-1 x + log_e sqrt((1+x)/(1-x)) - 2x) / x^5 = 2/5 Statement II : lim x->1 (x^(2/(1-x))) = 1/e^2 In the light of the above statements, choose the correct answer from the options given below

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q16 – JEE-MAIN 2025

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Home Maths Limits and DerivativesQuestion #279

Given below are two statements: Statement I :

lim_{x \to 0} (\frac{\tan^{- 1} x + \log_{e} \sqrt{\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

Clue 1 of 3: Limit Evaluation Techniques

For limits involving indeterminate forms like $0 / 0$ or $1^{\infty}$ , consider using L'Hopital's rule or Taylor series expansions.

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Step 1: Evaluate Statement I using Taylor Series✦ Active

First, simplify the logarithmic term: $\log_{e} \sqrt{\frac{1 + x}{1 - x}} = \frac{1}{2} (\log_{e} (1 + x) - \log_{e} (1 - x))$ . Now, use the Maclaurin series expansions for each term around $x = 0$ up to $x^{5}$ :

\tan^{- 1} x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - O (x^{7})

\log_{e} (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \frac{x^{5}}{5} - O (x^{6})

\log_{e} (1 - x) = - x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - \frac{x^{5}}{5} - O (x^{6})

\frac{1}{2} (\log_{e} (1 + x) - \log_{e} (1 - x)) = \frac{1}{2} ((x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \frac{x^{5}}{5}) - (- x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - \frac{x^{5}}{5}))

= \frac{1}{2} (2 x + \frac{2 x^{3}}{3} + \frac{2 x^{5}}{5} + O (x^{6})) = x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + O (x^{6})

Substitute these into the numerator of Statement I:

(x - \frac{x^{3}}{3} + \frac{x^{5}}{5}) + (x + \frac{x^{3}}{3} + \frac{x^{5}}{5}) - 2 x + O (x^{6})

= (x + x - 2 x) + (- \frac{x^{3}}{3} + \frac{x^{3}}{3}) + (\frac{x^{5}}{5} + \frac{x^{5}}{5}) + O (x^{6})

= 0 + 0 + \frac{2 x^{5}}{5} + O (x^{6})

Therefore, the limit is:

lim_{x \to 0} \frac{\frac{2 x^{5}}{5} + O (x^{6})}{x^{5}} = \frac{2}{5}

Statement I is true.

Step 2: Evaluate Statement II using the

e^{g (x) (f (x) - 1)}

form○ Expand

The limit in Statement II is of the indeterminate form $1^{\infty}$ . We use the property $lim_{x \to a} [f (x)]^{g (x)} = e^{lim_{x \to a} g (x) [f (x) - 1]}$ . Here, $f (x) = x$ and $g (x) = \frac{2}{1 - x}$ .

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

Evaluate the exponent limit:

lim_{x \to 1} (\frac{2}{1 - x} \cdot (x - 1)) = lim_{x \to 1} (\frac{2}{- (x - 1)} \cdot (x - 1)) = lim_{x \to 1} (- 2) = - 2

So, the limit is $e^{- 2} = \frac{1}{e^{2}}$ . Statement II is true.

Step 3: Conclusion○ Expand

Since both Statement I and Statement II are true, the correct option is 1.

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Maths Question 16 – JEE-MAIN 2025

Clue 1 of 3: Limit Evaluation Techniques

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