# 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.

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

• So I need to ensure that numerator and denominator have the same $+ o(...)$ ? – rehband Aug 12 '14 at 15:35
• Why you want to have the same $o(\cdots)$? No in general it's not the case. – user63181 Aug 12 '14 at 15:41
• You can expand the fraction and then you get 1/120 + o(1), so 1/20 plus a term that tends to zero if x goes to zero. In general, you need to expand as far as needed to get the limit in the form of answer + o(x^n) with n >= 0. – Count Iblis Aug 12 '14 at 15:42
• You're welcome @rehband and good luck in your exam:-) – user63181 Aug 12 '14 at 15:46
• Thanks. Still a month to go, I'll ask plenty more questions before :) – rehband Aug 12 '14 at 15:55

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 $x\to 0^-$.)

• I just let my post then you can see the mistakes (and not reproduce). I made $6\sinh x-6x-x^3=6x-6x-x^3+o(x)=-x^3+o(x)$ but this give $0+o(x)$ and the all function came nul. So you have to go far enough to not arrive in the case. As you can see, to develop until a $o(x^3)$ wouldn't be enough because we arrive to $0+o(x^3)$, then you have ton go at minimum to an $o(x^5)$ (because an $o(x^4)$ arrive to $0+o(x^4)$ too. So consider not my answer like an official answer but more like an exemple of what arrive if you don't go far enough (And I hope you will not reproduce it to your exam :-) – idm Aug 12 '14 at 15:50
• Ok, great, got it. Good to learn from mistakes too. Thank you! – rehband Aug 12 '14 at 15:58