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

$\begin{array}{l}\underset{\mathrm{x}\to 0}{lim} \frac{\cos (\tan \mathrm{x})-\cos \mathrm{x}}{{\mathrm{x}}^4}=\underset{\mathrm{x}\to 0}{lim} \frac{2\sin \left(\frac{\mathrm{x}+\tan \mathrm{x}}{2}\right)\sin \left(\frac{\mathrm{x}-\tan \mathrm{x}}{2}\right)}{{\mathrm{x}}^4}\\ =\underset{\mathrm{x}\to 0}{lim} \frac{2\sin \left(\frac{\mathrm{x}+\tan \mathrm{x}}{2}\right)\sin \left(\frac{\mathrm{x}-\tan \mathrm{x}}{2}\right)}{{\mathrm{x}}^4\left(\frac{\mathrm{x}+\tan \mathrm{x}}{2}\right)\left(\frac{\mathrm{x}-\tan \mathrm{x}}{2}\right)}\left(\frac{{\mathrm{x}}^2-{\tan }^2\mathrm{x}}{4}\right)\\ =\frac{1}{2}\underset{\mathrm{x}\to 0}{lim} \frac{{\mathrm{x}}^2-{\tan }^2\mathrm{x}}{{\mathrm{x}}^4}\\ =\frac{1}{2}\underset{\mathrm{x}\to 0}{lim} \frac{{\mathrm{x}}^2-{\left(\mathrm{x}+\frac{{\mathrm{x}}^3}{3}+\frac{2}{15}{\mathrm{x}}^4+\dots \right)}^2}{{\mathrm{x}}^4}\\ =\frac{1}{2}\underset{\mathrm{x}\to 0}{lim} \frac{1}{{\mathrm{x}}^2}\left(1-{\left(1+\frac{{\mathrm{x}}^2}{3}+\frac{2}{15}{\mathrm{x}}^4+\dots \right)}^2\right)=-\frac{1}{3}\end{array}$