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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

  1. A \(\frac{1}{2}\)
  2. B \(\frac{1}{3}\)
  3. C \(\frac{1}{4}\)
  4. D \(\frac{1}{6}\)
Verified Solution

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 \(x \rightarrow 0^{+}\)…