$$\begin{array}{cl}
& \displaystyle \lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{xe^x + xe^{-x} - 2x\cos x}{e^x - e^{-x} - 2\sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{x + x^2 + \frac12x^3 + o(x^4) + x - x^2 + \frac12x^3 + o(x^4) - 2x + x^3 + o(x^4)} {1 + x + \frac12x^2 + \frac16x^3 + o(x^4) - 1 + x - \frac12x^2 + \frac16x^3 + o(x^4) - 2x + \frac13x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{2x^3 + o(x^4)} {\frac23x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{3 + o(x)} {1 + o(x)} \\
=& 3
\end{array}$$
The denominator is of the fifth degree (after linearizing the tangent), so if there is a finite answer you will need five successive applications of L'Hospital.
The numerator is easy:
$$\sinh x-x\cosh x+\frac{x^3}3,$$
$$-x\sinh x+x^2,$$
$$-\sinh x-x\cosh x+2x,$$
$$-2\cosh x-x\sinh x+2$$
$$-3\sinh x-x\cosh x,$$
$$-4\cosh x-x\sinh x.$$
Every time, you need to check that the expression tends to zero (otherwise the limit will not exist because of the zero denominator).
For the denominator, it is really worth to rewrite
$$x^2\tan^3x=x^5\frac{\tan^3x}{x^3}$$ and take the fraction away. Then the fifth derivative is $5!$ and the requested ratio
$$-\frac1{30}.$$
For info, keeping the denominator as is, we get
$$2520x^2(\tan(x))^8+6600x^2(\tan(x))^6+3600x(\tan(x))^7+36x^2(\sec(x))^2(\tan(x))^2+\\
5772x^2(\tan(x))^4+8160x(\tan(x))^5+1200(\tan(x))^6+120x\tan(x)(\sec(x))^2+
\\120x^2(\sec(x))^2+1692x^2(\tan(x))^2+5640x(\tan(x))^3+2280(\tan(x))^4+
\\1080x\tan(x)+120(\sec(x))^2+1080(\tan(x))^2$$
and the only nonzero term is $120\sec^2x$.
Best Answer
The first and second derivatives of both numerator and denominator are zero. The third derivative of the numerator at $x=0$ is $3.$ The third derivative of the denominator at $x=0$ is $-3.$
The limit is $-1.$