What's the implication of l'Hospital's rule on rate of convergence?

Question

Consider $h(x)=f(x)/g(x)$, if l'Hospital's rule is applicable, then $$\lim h(x)=\lim\frac{f'(x)}{g'(x)}.$$ Does this fact implies $h(x)$ and $f'(x)/g'(x)$ converge at the same speed? E.g. if $f'(x)/g'(x)\to L$ linearly, can we say the same about $h(x)$? If not, can we say anything in general about the rate of convergence?

Answer 1

In fact, as long as the functions in question are analytic in a neighborhood of the limit point, the rate of convergence should be the same. To see this, develop both numerator and denominator as Taylor series; for concreteness' sake, we'll take the limit as $\lim\limits_{x\to0}$. Then the numerator will be of the form $Ax^n+\Theta(x^{n+a})$ for some positive integer $a$, and the denominator will be of the form $Bx^n+O(x^{n+1})$, yielding an expression of the form $\frac AB+\Theta(x^a)$; in other words, the convergence will be as some polynomial in $x$ (typically linear, but not always — for instance, $\frac{\cos x-1}{x^2}$ converges quadratically to $\frac12$). Moreover, the degree of this polynomial will stay the same under the 'L'Hôpital operation', because the numerator becomes $Anx^{n-1}+\Theta(x^{n+a-1})$ while the denominator becomes $Bnx^{n-1}+O(x^n)$, and the quotient of these is still of the form $\frac AB+\Theta(x^a)$.

Answer 2

Barry Cipra's counter-example is very nice. If $\forall x \in \mathbb{R}, \begin{cases} f(x)= x^2 \\ g(x)=e^x \end{cases}$

\begin{equation} \lim_{x \to \infty} \frac{x^2}{e^x}=\lim_{x \to \infty} \frac{2x}{e^x} \end{equation}

Yet the two expressions don't approach $0$ at the same rate since:

\begin{equation} \lim_{x \to \infty} \frac{x^2/e^x}{2x/e^x}=\infty \end{equation}

For this reason, this question is best handled on a case by case basis unless you have prior knowledge that $f$ and $g$ belong to a certain asymptotic class. Ex: polynomial, exponential.