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We are given two functions, namely: $f, g: U^\circ\subset \mathbb{R}\to \mathbb{R}$ where $0 \in \overline{U}$.

Suppose you want to compute the limit $$\lim_{x \to 0} \frac{f(x)}{g(x)}$$ and that $\lim_{x \to 0} f(x) = \lim_{x\to 0} g(x) = 0$. In addition, suppose that l'Hopital's theorem hypothesis are satisfied and that exploiting l'Hopital rule we found that $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = L \in \{\mathbb{R}, \pm\infty\}.$$ Can the same result be obtained using Taylor's approximation? And if we consider instead $\lim_{x \to +\infty} \frac{f(x)}{g(x)}$ anything changes?

My speculations and question's motivation follow.

In order to apply Taylor's approximation one has to choose a function, an approximation point and an order of approximation. Taylor's formula gives us two objects: a polynomial and a reminder. But the only thing we know about the reminder is its behavior near the approximation point. So in order to apply Taylor to the functions $f,g$ we have to choose 0 as approximation point (otherwise how can we get rid of the reminder?). It follows that $f,g$ have to be defined in $0$. This can be problematic, for example if $f(x) = \log(x) $ and $g(x) = x^{-1}$ we can apply l'Hopital's rule and get to $$ \frac{-x^2}{x} \to 0.$$ But we are unable to expand $f$ or $g$ near $0$ as them are not defined in $0$.

So it seems to me that Taylor method alone cannot cover all cases of interest. But I must be wrong since I've read Why are people so interested in finding limits without l'Hôpital's rule? and I figured out that many people rely only on Taylor's method (which is a beautiful thing in my opinion) and that in some universities L'Hopital's rule use is even discouraged.

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    $\begingroup$ Does differentiability imply existence of the Taylor series? $\endgroup$ – Git Gud Feb 25 '15 at 19:07
  • $\begingroup$ We can use for sure a first order approximation of $f,g$ in an open interval with endpoint $0$. But it would be interesting also to consider the case where $f$ and $g$ are defined and differentiable at $0$ or are analytic. $\endgroup$ – Warlock of Firetop Mountain Feb 25 '15 at 19:20
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    $\begingroup$ L'Hospital's rule is equivalent to the use of first order approximations — which are more commendable than L'Hospital's rule $\endgroup$ – Bernard Feb 25 '15 at 20:10
  • $\begingroup$ This is exactly the statement I was searching for, where can I find the proof? $\endgroup$ – Warlock of Firetop Mountain Feb 25 '15 at 21:04

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