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

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?

• Do $x^2/e^x$ and $2x/e^x$ converge (to $0$, as $x\to\infty$) at the same rate? Commented Mar 25, 2016 at 14:26
• What is your exact definition of "speed/rate of convergence"? Commented Mar 25, 2016 at 14:27
• @DanielRobert-Nicoud: As in this Commented Mar 25, 2016 at 14:38
• Does L'Hôpital's rule actually apply to $x^2/e^x$? It's not an indeterminate form... Commented Oct 19, 2017 at 22:41

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 $\frac AB+\Theta(x^a)$.

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

$$\lim_{x \to \infty} \frac{x^2}{e^x}=\lim_{x \to \infty} \frac{2x}{e^x}$$

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

$$\lim_{x \to \infty} \frac{x^2/e^x}{2x/e^x}=\infty$$

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.

• As I just commented on the main Q, I believe this example is invalid because it doesn't meet the applicability conditions for applying L'Hôpital's rule (in particular, it's not an indeterminate form). Commented Oct 19, 2017 at 22:46
• $\lim_{x\to\infty} x^2/e^x$ is indeed an indeterminate form. Commented Oct 20, 2017 at 0:30
• Ahh, fair - for some reason I was thinking it was $\lim_{x\to 0}$, mea culpa. Commented Oct 20, 2017 at 7:54