A calculus problem by Akshat Sharda

By Maclaurin series: $\begin{aligned} \cos(\sin x) &= 1 - \frac{1}{2}\sin^2x + \frac{1}{24}\sin^4 x + o(x^4) \\ &= 1 - \frac{1}{2}\left(x - \frac{1}{6}x^3 + o(x^3)\right)^2 + \frac{1}{24}\big(x + o(x)\big)^4 + o(x^4) \\ &= 1 - \frac{1}{2}\left(x - \frac{1}{6}x^3\right)^2 + \frac{1}{24}x^4 + o(x^4) \\ &= 1 - \frac{1}{2}x^2 + \frac{1}{6}x^4 + \frac{1}{24}x^4 + o(x^4).\end{aligned}$

Then subtracting $\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4)$ yields $\cos(\sin x) - \cos x = \frac{1}{6}x^4 + o(x^4).$

Therefore, $\lim_{x\rightarrow 0} \frac{\cos(\sin x) - \cos x}{x^4} = \lim_{x\rightarrow 0} \frac{\tfrac{1}{6}x^4 + o(x^4)}{x^4} = \lim_{x\rightarrow 0} \frac{1}{6} + o(1) = \frac{1}{6}.$