Valid Use of l’Hospital’s Rule and Recursive Application

calculuslimits

$$\lim_{x \to 0} \frac{1}{x} + \frac{1}{x}$$
$$\lim_{x \to 0} \frac{2x}{x^2}$$

Since this evaluates to an indeterminate form $\frac{0}{0}$ we use l'Hospital's Rule:

$$\lim_{x \to 0} \frac{2}{2x}$$

Since this also evaluates to an indeterminate form $\frac{2}{0}$ we use l'Hospital's Rule again:

$$\lim_{x \to 0} \frac{0}{2}=0$$

I know that I could have simply divided both numerator and denominator by $x$ to get the same result. This is just an example to ask the question: Can l'Hospital's Rule be used recursively?

EDIT: Sorry, I messed up my example. I cannot think of a good example right now but the question still stands. Can one use l'Hospital's Rule recursively?

Best Answer

L'Hôpital's Rule

Assuming that the following conditions are true:

  1. $f(x)$ and $g(x)$ must be differentiable
  2. $\frac{d}{dx}g(x)\neq 0$
  3. $\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{0}{0}\mbox{ or }\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{\pm\infty}{\pm\infty}$

Then, $$ \lim\limits_{x\to c} \frac{f(x)}{g(x)}= \lim\limits_{x\to c} \frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)}=L $$ Where $c$ and $L$ is any real number or $\pm\infty$.

So to answer your questions, yes, L'Hôpital's rule can be used repeatedly, provided that all of the above conditions are met. Since your example doesn't meet the aforementioned conditions, L'Hôpital's rule is not applicable.

Here is a case where L'Hôpital's rule is applicable multiple times, $$ \lim_{x\to \infty}\dfrac{e^x}{x^2}=\frac{\infty}{\infty} $$ Since the conditions are met, we can apply L'Hôpital's rule $$ \lim_{x\to \infty} \frac{\frac{d}{dx}e^x}{\frac{d}{dx}x^2}= \lim_{x\to \infty} \frac{e^x}{2x}= \frac{\infty}{\infty} $$ Notice that the conditions are met again, so now $$ \lim_{x\to \infty} \frac{\frac{d}{dx}e^x}{\frac{d}{dx}2x}= \lim_{x\to \infty} \frac{e^x}{2}= \infty $$ Therefore $$ \lim_{x\to \infty}\frac{e^x}{x^2}=\infty $$

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