Is it possible to evaluate $\lim_{x\to 0}\frac{\sinh x-x\cosh x+\frac{x^3}3}{x^2\tan^3x}$ applying L'Hopital's Rule? I evaluated following limit with taylor series but for a practice I am trying to evaluate it using L'Hopital's Rule:
$$\lim_{x\to 0}\frac{\sinh x-x\cosh x+\frac{x^3}3}{x^2\tan^3x}=\lim_{x\to0}\cfrac{f(x)}{g(x)}$$
$f(x)=\sinh x-x\cosh x+\frac{x^3}3  ,f(0)=0$
$f'(x)=-x\sinh x+x^2, f'(0)=0$
$f''(x)=-\sinh x-x\cosh x+2x, f''(0)=0$
$f'''(x)=-2\cosh x-x\sinh x+2, f'''(0)=0$
It seems it is going to be $0$ for further derivatives.
Also for $g(x)=x^2\tan^3x$, wolfram alpha gives this result:

Which seems we have $g^{(n)}(x)=0$ too.
So Is there any way to evaluate the limit applying L'Hopital's Rule?
 A: 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 $120\sec^2x$.
A: To simplify the evaluation we can use that
$$\frac{\sinh x-x\cosh x+\frac{x^3}3}{x^2\tan^3x}=\frac{x^3}{\tan^3x}\frac{\sinh x-x\cosh x+\frac{x^3}3}{x^5}$$
and since $\frac{x^3}{\tan^3x} \to 1$ we reduce to evaluate by l'Hospital
$$\lim_{x\to 0}\frac{\sinh x-x\cosh x+\frac{x^3}3}{x^5}$$
