L'Hôpital's Rule
A Systematic Derivative-Based Tool for Evaluating Indeterminate Limits
1. The Setting Recalled: Indeterminate Forms
In our earlier study of limits, we encountered situations where direct substitution produced expressions whose values could not be determined without further analysis. These are the indeterminate forms:
Back then, we handled these with algebraic tricks: factoring, rationalizing (conjugate multiplication), cancellation, and known remarkable limits such as:
$$ \lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}, \qquad \lim_{x \to 0} \frac{e^x - 1}{x} = 1 $$Those methods work beautifully — but they require ingenuity for each new problem. L'Hôpital's Rule fulfils the promise of a systematic, derivative-based tool that handles the two fundamental indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$ directly, and to which all other indeterminate forms can be reduced.
2. Statement of L'Hôpital's Rule
2.1 The $\frac{0}{0}$ Form
Theorem (L'Hôpital's Rule — Zero-over-Zero).
Let $f$ and $g$ be functions differentiable on an open interval $I$ containing $c$ (except possibly at $c$ itself). Suppose:
- $\displaystyle\lim_{x \to c} f(x) = 0$ and $\displaystyle\lim_{x \to c} g(x) = 0$;
- $g'(x) \neq 0$ for all $x \in I$ with $x \neq c$;
- $\displaystyle\lim_{x \to c} \frac{f'(x)}{g'(x)}$ exists (or equals $\pm\infty$).
Then:
$$ \boxed{\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}} $$Applicability: The rule applies at finite points ($c \in \mathbb{R}$), for one-sided limits ($x \to c^+$ or $x \to c^-$), and at infinity ($x \to +\infty$ or $x \to -\infty$).
2.2 The $\frac{\infty}{\infty}$ Form
Theorem (L'Hôpital's Rule — Infinity-over-Infinity).
If $\displaystyle\lim_{x \to c} f(x) = \pm\infty$ and $\displaystyle\lim_{x \to c} g(x) = \pm\infty$, and the other hypotheses above hold, then the same conclusion follows:
$$ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} $$2.3 The Engine Behind the Proof: Cauchy's Mean Value Theorem
Cauchy's Mean Value Theorem. Suppose $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, and $g'(x) \neq 0$ on $(a,b)$. Then there exists at least one $c \in (a,b)$ such that:
$$ \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} $$Note: When $g(x) = x$, this reduces to the ordinary Mean Value Theorem.
Proof Sketch (0/0 case, right-hand limit)
We sketch the proof for $\lim_{x \to a^+} \frac{f(x)}{g(x)}$ where $f(a) = g(a) = 0$:
- Since $g'(x) \neq 0$ near $a$, by the Mean Value Theorem, $g(x) \neq 0$ near $a$ (otherwise Rolle's theorem would force $g'$ to vanish).
- For any $x$ near $a$ (with $x > a$), apply Cauchy's MVT on $[a, x]$: there exists $c_x \in (a, x)$ with $$ \frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(c_x)}{g'(c_x)} $$
- As $x \to a^+$, we have $c_x \to a^+$ (since $a < c_x < x$), so $$ \lim_{x \to a^+} \frac{f(x)}{g(x)} = \lim_{c_x \to a^+} \frac{f'(c_x)}{g'(c_x)} = \lim_{x \to a^+} \frac{f'(x)}{g'(x)} $$
The $\frac{\infty}{\infty}$ case requires a more delicate argument (rewriting the ratio using Cauchy's MVT on intervals $[x, x_1]$ and controlling remainder terms as $g(x) \to \infty$), but the engine is the same theorem.