Given below are two statements: Statement I: \lim\limits_x \to 0 ( \frac\tan^-1 x + \log_e \sqrt\frac1+x1-x - 2xx^5 ) = \frac25 Statement II: \lim\limits_x \to 1 ( x^\frac21-x ) = \frac1e^2 In the light of the above statements, choose the correct answer from the options given below:

Both Statement I and Statement II are true

Statement I is false but Statement II is true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Limits Continuity and Differentiability PYQ — JEE Main 2025 · 8 Apr · Shift 2 · Q36 with Solution

View written solutionFree

Correct answer: C

Check Statement I

We need to evaluate

L=\lim_{x\to 0}\frac{\tan^{-1}x+\log_e\sqrt{\frac{1+x}{1-x}}-2x}{x^5}.

First simplify the logarithmic term:

\log_e\sqrt{\frac{1+x}{1-x}}=\frac12\log\left(\frac{1+x}{1-x}\right).

Using the standard expansion,

\frac12\log\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots

Also,

\tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots

Now add:

\tan^{-1}x+\frac12\log\left(\frac{1+x}{1-x}\right) =\left(x-\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)+\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right).

So,

\tan^{-1}x+\log_e\sqrt{\frac{1+x}{1-x}}=2x+\frac{2x^5}{5}+\cdots

Hence,

\tan^{-1}x+\log_e\sqrt{\frac{1+x}{1-x}}-2x=\frac{2x^5}{5}+\cdots

Therefore,

L=\lim_{x\to 0}\frac{\frac{2x^5}{5}+\cdots}{x^5}=\frac25.

So Statement I is true.

Check Statement II

We need to evaluate

\lim_{x\to 1}x^{\frac{2}{1-x}}.

This is of the form $1^\infty$ , so take logarithm.

Let

y=x^{\frac{2}{1-x}}.

Then

\ln y=\frac{2\ln x}{1-x}.

Now evaluate

\lim_{x\to 1}\frac{2\ln x}{1-x}=2\lim_{x\to 1}\frac{\ln x}{1-x}.

Using L'Hospital's Rule:

\lim_{x\to 1}\frac{\ln x}{1-x} =\lim_{x\to 1}\frac{1/x}{-1}=-1.

Thus,

\lim_{x\to 1}\ln y=2(-1)=-2.

So,

\lim_{x\to 1}y=e^{-2}=\frac{1}{e^2}.

Hence Statement II is true.

Conclusion

Statement I is true
Statement II is true

Therefore, the correct option is

\boxed{\text{C}}

Comparison with stored answer

Stored correct answer: C

This matches our derived answer.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: C

Check Statement I

We need to evaluate

L=\lim_{x\to 0}\frac{\tan^{-1}x+\log_e\sqrt{\frac{1+x}{1-x}}-2x}{x^5}.

First simplify the logarithmic term:

\log_e\sqrt{\frac{1+x}{1-x}}=\frac12\log\left(\frac{1+x}{1-x}\right).

Using the standard expansion,

\frac12\log\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots

Also,

\tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots

Now add:

\tan^{-1}x+\frac12\log\left(\frac{1+x}{1-x}\right) =\left(x-\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)+\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right).

So,

\tan^{-1}x+\log_e\sqrt{\frac{1+x}{1-x}}=2x+\frac{2x^5}{5}+\cdots

Hence,

\tan^{-1}x+\log_e\sqrt{\frac{1+x}{1-x}}-2x=\frac{2x^5}{5}+\cdots

Therefore,

L=\lim_{x\to 0}\frac{\frac{2x^5}{5}+\cdots}{x^5}=\frac25.

So Statement I is true.

Check Statement II

We need to evaluate

\lim_{x\to 1}x^{\frac{2}{1-x}}.

This is of the form $1^\infty$ , so take logarithm.

Let

y=x^{\frac{2}{1-x}}.

Then

\ln y=\frac{2\ln x}{1-x}.

Now evaluate

\lim_{x\to 1}\frac{2\ln x}{1-x}=2\lim_{x\to 1}\frac{\ln x}{1-x}.

Using L'Hospital's Rule:

\lim_{x\to 1}\frac{\ln x}{1-x} =\lim_{x\to 1}\frac{1/x}{-1}=-1.

Thus,

\lim_{x\to 1}\ln y=2(-1)=-2.

So,

\lim_{x\to 1}y=e^{-2}=\frac{1}{e^2}.

Hence Statement II is true.

Conclusion

Statement I is true
Statement II is true

Therefore, the correct option is

\boxed{\text{C}}

Comparison with stored answer

Stored correct answer: C

This matches our derived answer.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

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