JEE Mains · Maths · STD 11 - 12. limits
Given below are two statements :
Statement I : \(\lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x+\log _e \sqrt{\frac{1+x}{1-x}}-2 x}{x^5}\right)=\frac{2}{5}\)
Statement II : \(\lim _{\mathrm{x} \rightarrow 1}\left(\mathrm{x}^{\frac{2}{1-\mathrm{x}}}\right)=\frac{1}{\mathrm{e}^2}\)
In the light of the above statements, choose the correct answer from the options given below :
- A Statement I is false but Statement II is true
- B Statement I is true but Statement II is false
- C Both Statement I and Statement II are false
- D Both Statement I and Statement II are true
Answer & Solution
Correct Answer
(D) Both Statement I and Statement II are true
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[\ln (1+x)-\ln (1-x)]-2 x}{x^5} \)…
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