calculus - Evaluating a limit with Taylor Series

I would like to find the following limit using Taylor Series:

$$\lim_{x\to0}\frac{6\sinh x-6x-x^3}{x^4(6+x^2)\sinh x}.$$

Now my question is the following: How do I know exactly how to approximate the numerator and the denominator? We're clearly going to need at least the first three terms of $\sinh x = x + \frac{x^3}{6} + \frac{x^5}{120}+...$ in the numerator to get something nonzero.

Could you please show me what a rigorous application of Taylors theorem would look like in this case?

(This is not homework - I'm preparing for an exam)

Thank you.

Answer

You ask for rigorous application? What about this:

$$\lim_{x\to0}\frac{6\sinh x-6x-x^3}{x^4(6+x^2)\sinh x}=\lim_{x\to0}\frac{6\left(x + \frac{x^3}{6} + \frac{x^5}{120}+o(x^5)\right)-6x-x^3}{x^4(6+x^2)( x+o(x))}\\=\lim_{x\to0}\frac{\frac{x^5}{20}+o(x^5)}{6x^5+o(x^5)}=\frac1{120}\quad?$$