[Math] Using Taylor Expansion to evaluate limits

Question

I am going through some lecture notes and I came across this limit:

$$\lim_{x\to 0}\frac{\sinh x^4-x^4}{(x-\sin x)^4} $$

In the notes, it says (after introducing L'Hopital's Rule) that this would be difficult to evaluate using L'Hopital's Rule but can be done on sight using Taylor's Theorem. After reading the section on Taylor's Theorem, I don't understand how this can be done in sight.

Would one need to calculate its Taylor expansion? If so, how would one go about doing that as its derivatives aren't defined at 0? I have used Wolfram to see the Taylor expansion is $216+O(x^2)$ which means the limit is equal to 216, but how does one calculate this Taylor expansion?

Answer 1

Hint Expanding we have that $$\sinh x^4 = x^4 + \frac{(x^4)^3}{3!} + O(x^{13}) ,$$ so the numerator is $$\sinh x^4 - x^4 = \frac{x^{12}}{3!} + O(x^{13}).$$ On the other hand, $$\sin x = x - \frac{x^3}{3!} + O(x^4) ,$$ so the denominator is $$(x - \sin x)^4 = \left(\frac{x^3}{3!} + O(x^4)\right)^4 = \cdots$$

$$ \cdots = \left(\frac{x^3}{3!}\right)^4 + O(x^{13}) = \frac{x^{12}}{(3!)^4} + O(x^{13}).$$ The leading term of the quotient is the quotient of the leading terms, namely, $$\frac{(3!)^4}{3!} = 3!^3 = 216 .$$ The fact that the leading terms are both comparable to $x^{12}$ tells us that we would need to apply l'Hopital's Rule 12 times before being able to evaluate---needless to say, this a much faster method.