$\underset{x→0}{\lim}\frac{\cos(\tan x)-\cos x}{x^4}$ is equal to:

$-\frac{1}{3}$

cos (tan x) – cos x = $2\sin(\frac{x+\tan x}{2}).\sin(\frac{x-\tan x}{2})$

$L=\underset{x→0}{\lim}\frac{2\sin(\frac{x+\tan x}{2}).\sin(\frac{x-\tan x}{2})}{x^4}=\frac{1}{2}\underset{x→0}{\lim}\frac{x^2-\tan^2x}{x^4}=\frac{x^2-(x+\frac{x^3}{3}+\frac{2}{15}x^5+...)^2}{x^4}$

$L=\frac{1}{2}\underset{x→0}{\lim}\frac{1}{x^2}(1-(1+\frac{x^2}{3}+...)^2)=\frac{1}{2}\underset{x→0}{\lim}\frac{(2+\frac{x^2}{3}+\frac{2}{15}x^4...)(\frac{-x^2}{3}-\frac{2}{15}x^4...)}{x^2}$

$=\frac{1}{2}\underset{x→0}{\lim}(→2)(\frac{-1}{3}-\frac{2}{15}x^2...)=\frac{1}{2}\underset{x→0}{\lim}(→2)(→\frac{-1}{3})=\frac{-1}{3}$