Real Analysis 42 | L'Hospital's Rule [dark version]

TL;DR

The extended mean value theorem replaces the usual slope with (f(b)−f(a))/(g(b)−g(a)) and matches it to f′(x̂)/g′(x̂) at some interior point x̂.

Briefing Cornell Notes

Briefing

L’Hospital’s Rule is built from a more general “extended mean value theorem” that links two differentiable functions through a special mean slope. The key move is to replace the usual single-function mean slope with a ratio that combines both functions’ derivatives: for differentiable f and g on an interval, with g′ never zero inside the interval, there exists an interior point x̂ where

(f(b)−f(a))/(g(b)−g(a)) = f′(x̂)/g′(x̂).

That statement matters because it turns an otherwise hard limit of the form f(x)/g(x) into something computable from derivatives—provided the right hypotheses hold. The derivation uses the same backbone as the standard mean value theorem: construct an auxiliary function H that vanishes at the endpoints, then apply Rolle’s theorem to force H′(x̂)=0 at some interior point. The “extended” part comes from how H is defined: it subtracts a secant-like expression built from the quotient (f(b)−f(a))/(g(b)−g(a)), so differentiating H produces exactly the combined slope f′(x̂)/g′(x̂).

Once the extended mean value theorem is in place, L’Hospital’s Rule follows for limits that initially look like indeterminate forms. Consider a point x0 in an interval I and two differentiable functions f and g. The rule targets limits of the quotient f(x)/g(x) as x→x0, focusing on the common problematic case where f(x0)=0 and g(x0)=0. Under additional conditions—most importantly that g′ does not vanish near x0 except possibly at x0, and that the derivative quotient f′(x)/g′(x) has a (one-sided) limit—the theorem guarantees that the original quotient f(x)/g(x) has a matching limit.

The proof strategy is sequence-based: take any sequence x_n→x0 with x_n≠x0, apply the extended mean value theorem on the interval between x_n and x0, and obtain points x̂_n where the derivative quotient equals a secant quotient. Because f(x0)=g(x0)=0, the secant quotient collapses to f(x_n)/g(x_n). If the derivative quotient f′(x̂_n)/g′(x̂_n) converges (by assumption), then the secant quotient—and thus f(x_n)/g(x_n)—must converge to the same value. That forces the limit of f(x)/g(x) to exist and equal the limit of f′(x)/g′(x).

Two worked examples show the mechanics. For

lim_{x→0} log(1+x)/x,

the indeterminate form 0/0 is resolved by differentiating: (1/(1+x))/1 → 1. For

lim_{x→0} (1−cos x)/x,

differentiation gives sin(x)/2x, still 0/0, so the rule is applied again: cos(x)/2 → 1/2. In both cases, the derivative quotient’s limit determines the original limit, illustrating why the extended mean value theorem is the engine behind L’Hospital’s Rule.

Cornell Notes

L’Hospital’s Rule is justified using an extended mean value theorem that compares two functions f and g. If f and g are differentiable on an interval and g′ never equals 0 inside the interval, then there is an interior point x̂ where (f(b)−f(a))/(g(b)−g(a)) = f′(x̂)/g′(x̂). Rolle’s theorem drives the proof by applying it to a carefully constructed auxiliary function H that vanishes at the endpoints. With this tool, limits of the form f(x)/g(x) as x→x0 (especially when f(x0)=g(x0)=0) can be reduced to limits of f′(x)/g′(x), assuming g′ stays nonzero near x0 and the derivative quotient limit exists. Sequence arguments then force the original quotient limit to exist and match the derivative quotient’s value.

What changes when moving from the standard mean value theorem to the extended mean value theorem?

Instead of one function f producing a slope (f(b)−f(a))/(b−a), two functions f and g are used. The “mean slope” becomes a combined quotient (f(b)−f(a))/(g(b)−g(a)), and the interior-point conclusion matches it to a derivative ratio f′(x̂)/g′(x̂). A crucial assumption is that g′ is never 0 inside the interval, so the derivative ratio is well-defined and the secant quotient behaves properly.

How does the proof using Rolle’s theorem work in the extended setting?

An auxiliary function H is built so that H(a)=0 and H(b)=0. H is defined by subtracting from f(x) a secant-like expression whose denominator uses g(x)−g(a) and whose numerator uses the corresponding endpoint values of f and g. When H′(x̂)=0 at some interior x̂ (guaranteed by Rolle’s theorem), differentiating H yields the identity f′(x̂) − [(f(b)−f(a))/(g(b)−g(a))]·g′(x̂)=0, which rearranges to (f(b)−f(a))/(g(b)−g(a)) = f′(x̂)/g′(x̂).

Why does L’Hospital’s Rule require f(x0)=0 and g(x0)=0?

The rule targets indeterminate forms, especially 0/0. If f(x0) and g(x0) are both zero, then the quotient f(x)/g(x) may still have a meaningful limit even though direct substitution gives an undefined expression. In the proof, this condition makes the secant quotient collapse: when applying the extended mean value theorem on intervals between x_n and x0, the endpoint terms at x0 vanish, leaving exactly f(x_n)/g(x_n).

What role does the condition “g′ does not vanish near x0” play?

It ensures the derivative quotient f′(x)/g′(x) is meaningful and that the extended mean value theorem can be applied without dividing by zero. The transcript frames it as g′ being nonzero in a neighborhood around x0 except possibly at x0 itself. That prevents the interior-point ratio from becoming undefined and supports the existence of the derivative-limit that drives the conclusion.

How does the sequence argument force the limit of f(x)/g(x) to exist?

Choose any sequence x_n→x0 with x_n≠x0. Apply the extended mean value theorem on the interval between x_n and x0 to obtain points x̂_n where f(x_n)/g(x_n) equals f′(x̂_n)/g′(x̂_n). If the derivative quotient has a limit L, then the right-hand side converges to L, so the left-hand side converges to L for every such sequence. That forces the function limit f(x)/g(x) as x→x0 to exist and equal L.

Why do the example limits work after one or two applications of L’Hospital’s Rule?

For log(1+x)/x, differentiating gives (1/(1+x))/1, whose limit at 0 is 1, so the original limit is 1. For (1−cos x)/x, the first differentiation yields sin(x)/(2x), which is still 0/0 at x=0; applying L’Hospital again gives cos(x)/2, whose limit at 0 is 1/2. The process stops once the derivative quotient has a definite limit.

Review Questions

State the extended mean value theorem and list the key assumptions on f and g.
In the L’Hospital proof, why does the condition f(x0)=g(x0)=0 simplify the secant quotient?
For lim_{x→0} (1−cos x)/x, what derivative quotient appears after the first and second applications of L’Hospital’s Rule?

Key Points

1
The extended mean value theorem replaces the usual slope with (f(b)−f(a))/(g(b)−g(a)) and matches it to f′(x̂)/g′(x̂) at some interior point x̂.
2
A nonvanishing condition on g′ inside the interval is essential to make the derivative ratio well-defined.
3
Rolle’s theorem is applied to an auxiliary function H engineered to vanish at the endpoints, forcing H′(x̂)=0.
4
L’Hospital’s Rule targets limits where f(x0)=0 and g(x0)=0, turning an indeterminate 0/0 into a derivative-based limit.
5
If the limit of f′(x)/g′(x) exists (one-sided as needed) and g′ stays nonzero near x0, then the limit of f(x)/g(x) exists and equals that derivative limit.
6
Sequence arguments show that convergence of the derivative quotient forces convergence of f(x)/g(x) for every sequence approaching x0.

Highlights

The extended mean value theorem is the engine behind L’Hospital’s Rule: (f(b)−f(a))/(g(b)−g(a)) equals f′(x̂)/g′(x̂) for some interior x̂.

Constructing H so it vanishes at endpoints lets Rolle’s theorem produce the exact derivative ratio needed.

For log(1+x)/x, one differentiation resolves the 0/0 form immediately, yielding a limit of 1.

For (1−cos x)/x, one application still leaves 0/0, but a second application gives cos(x)/2 and the limit 1/2.

Topics

Extended Mean Value Theorem
Rolle’s Theorem
L’Hospital’s Rule
Indeterminate Forms
Limit Computation