# Do you need L'Hôpital's rule to prove Taylor's formula?

$$\lim_{x\to0}\frac{\cos x-e^x}{\sin x}$$

without using L'Hôpital's rule. The answerer used the Taylor series expansion of the functions and proved that the limit equals $$-1$$. However, a comment (from an alleged Ph.D. in mathematics) stated that the answer wasn't correct, for you need L'Hôpital's rule to prove Taylor's formula, thus the answer was circular reasoning. Is that true? I don't remember seeing L'Hôpital's rule at all in the proofs I've read of the Taylor's formula.

• I'm not aware of any proof of Taylor's formula that uses (or needs to use) L'Hôpital's rule. Commented Jul 5 at 10:59
• Not at all necessary. Taylor’s theorem (in all its variants) holds in great generaloty even in Banach spaces (where we don’t really have a L’Hopital’s rule). The key is the mean-value inequality (in single variable calculus we have the mean value theorem, but that’s highly special to 1D, and also not necessary for most things). Commented Jul 5 at 11:04

The commenter may have been confused by the fact that give a Taylor polynomial $$P_{n, a}$$ of a function $$f$$ at the point $$a$$, then to prove that
$$\lim_{x \to a} \frac{f(x) - P_{n, a}(x)}{(x - a)^n} = 0$$
L'Hôpital's rule is usually used. And given that remainder term $$R_{n, a}$$ is defined as $$R_{n,a}= f - P_{n, a}$$, one can think that this property of $$P_{n, a}$$ may be essential, but in fact the proof for Taylor's theorem requires only the mean value theorem.