AP EAMCET · Maths · Limits
\(\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)}\) is equal to
- A \(\frac{1}{2}\)
- B \(\frac{1}{3}\)
- C \(\frac{1}{4}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\[ \lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \cdot \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)} \] \(\because\) We know that \(x \rightarrow 0^{+}\)…
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