How Does L'Hôpital's Rule Solve Indeterminate Forms in Calculus?

[Greg Bernhardt]

Definition/Summary

L'Hôpital's (or l'Hospital's) rule is a method for finding the limit of a function with an indeterminate form.

Equations

If the expression

$$\frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}$$

has the form $0/0$ or $\infty / \infty$, then l'Hôpital's rule states that

$$\lim_{x \rightarrow a} \frac{f(x)}{g(x)}<br /> = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$

provided that that second limit exists.

Extended explanation

Examples:

$$1.~~\lim_{x\rightarrow 0}\frac{\sin x}{x}\,=\,\lim_{x\rightarrow 0}\frac{\cos x}{1}\,=\,1$$

$$2.~~\lim_{x\rightarrow 0}\frac{e^x-1}{x}\,=\,\lim_{x\rightarrow 0}\frac{e^x}{1}\,=\,1$$

The rule can be applied more than once:

If after one application, the ratio is still of the form $0/0$ or $\infty / \infty$, then the rule may be applied again (and as many times as are needed to produce a limit):

$$3.~~\lim_{x\rightarrow 0}\frac{e^x\,-\,x\,-1}{\frac{1}{2}x^2}\,=\,\lim_{x\rightarrow 0}\frac{e^x\,-\,1}{x}\,=\,\lim_{x\rightarrow 0}\frac{e^x}{1}\,=\,1$$

Example of the rule not helping:

It is possible that the limit of the ratio of the derivatives does not exist, even though the limit of the original ratio does:

$$\lim_{x\rightarrow \infty}\frac{x\ +\ \sin x}{x}\ =\ 1$$

but $$\lim_{x\rightarrow \infty}\frac{1 +\ \cos x}{1}$$ does not exist.

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[fresh_42]

The exact wording is:

Be $I = ({\tilde{x}}_{0}, x_{0})$ a non-empty open interval and are $f, \, g \colon I \to \mathbb{R}$ differentiable functions for $x \nearrow x_{0}$ ($x$ goes from below against $x_{0}$) both converge to $0$ or both diverge definitely.

If $g'(x) \neq 0$ for all $x \in I$ holds and $\tfrac{f\,'(x)}{g' (x)}$ for $x \nearrow x_{0}$ converges against a value $q$ or definitely diverges, so does $\tfrac{f (x)}{g (x)}$. The same applies if we switch to $x \searrow x_{0}$ everywhere ($x$ goes from above against $\tilde{x}_{0}).$

Is $I$ a true subset of an open interval, on which the conditions are met, we have in particular
$$
\lim_{x \to x_{0}} \frac{f\,'(x)}{g' (x)} = q ~ \Longrightarrow ~ \lim_{x \to x_{0} } \frac{f (x)}{g (x)} = q
$$
The theorem also applies to improper interval limits $x_{0} = \pm \infty \,.$