Evaluating a limit with Taylor Series

Question

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 1

You ask for rigorous application? What about this:

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

Answer 2

There is not really a technique. Here you see that to just develop at the first degree, the indetermination is going to go because $\sinh(x)-6x=6x+o(x)=o(x)$ and the $x^3$ will be simplifie by the $x^4$, indeed, $\frac{x^3}{x^4}=\frac{1}{x}$.

Then $$\lim_{x\to 0}\frac{6\sinh(x)-6x-x^3}{x^4(6+x^2)\sinh x}=\lim_{x\to 0}\frac{6x-6x-x^3}{x^4(6+x^2)x}=\lim_{x\to 0}\frac{-1}{x(6+x^2)}$$ which doesn't exist. (because it gives $-\infty$ if $x\to 0^+$ and $+\infty$ if $x\to 0^- $.)