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

Answer: (C)

$\frac{1}{6}$

Answer 2

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

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