Question

\(\lim_{{x \to 0}} \limits\) \(\frac{cos(sin x)  -  cos x }{x^4}\) is equal to :

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{6}\)
• D. \(\frac{1}{12}\)

Answer 1

The Correct Option is C

The correct answer is (C) : \(\frac{-11}{9}\)

 \(\lim_{{x \to 0}} \limits\) \(\frac{cos(sinx) - cosx}{x^4}\) = \(\lim_{{x \to 0}} \limits\) \(\frac{2sin(x + sinx).sin(\frac{x - sinx}{2})}{x^4}\)
= \(\lim_{{x \to 0}} \limits\) \(2.\frac{(\frac{( x + sinx }{2})(\frac{x-sinx}{2})}{x^4}\)
= \(\lim_{{x \to 0}} \limits\) \(\frac{1}{2}.\) \((\frac{(x+x -\frac{x^3}{3!}+\frac{x^5}{5!}....)(x-x+\frac{x^3}{3!}....}{x^4})\)
= \(\lim_{{x \to 0}} \limits\) \(\frac{1}{2}.\) \((2-\frac{x^2}{3!}+\frac{x^4}{5!}....)(\frac{1}{3!}-\frac{x^2}{5!}-1)\)
= \(\frac{1}{6}\)