The Cauchy Mean Value Theorem

The Cauchy Mean Value Theorem is the load-bearing lemma behind L'Hopital's Rule: to prove that $\lim_{x\to a} f(x)/g(x) = \lim_{x\to a} f'(x)/g'(x)$ under indeterminate-form hypotheses, one applies the Cauchy Mean Value Theorem on shrinking intervals $[x, a]$ (or $[a,x]$) and lets $x \to a$, converting a ratio of function values into a ratio of derivative values at some intermediate point that gets squeezed toward $a$. Without the two-function comparison the Cauchy theorem provides, this passage from function ratios to derivative ratios has no rigorous justification. In Taylor's Theorem with Remainder, the Lagrange form of the remainder is derived by comparing the Taylor remainder function to an auxiliary power function via a generalized mean value argument that is, in essence, an iterated application of the same Rolle-based technique; some textbook derivations invoke the Cauchy Mean Value Theorem directly with $g(x) = (b-x)^{n+1}$ to produce the exact remainder formula. Beyond pure analysis, the theorem's underlying idea — comparing two quantities' rates of change via a common intermediate point — recurs in numerical analysis (error estimates for interpolation and quadrature rules), in the theory of convex functions (comparing difference quotients across sub-intervals), and in economics and physics whenever one needs to bound the ratio of accumulated changes in two coupled quantities by the ratio of their instantaneous rates at some common moment.

:::quiz[For $f(x)=x^3$, $g(x)=x^2+1$ on $[1,3]$, what is the value of $c$ guaranteed by the theorem?]

• $c = 2$
• $c = 13/6$ ✓
• $c = 7/3$
• $c = 5/2$

Ask the tutor: Can you walk through, step by step, how the Cauchy Mean Value Theorem is used inside the proof of L'Hopital's Rule?

Think of the pair $(g(t), f(t))$ as tracing a curve in the plane as $t$ runs from $a$ to $b$ — a parametric curve rather than a graph of a single function. The chord connecting the start point $(g(a), f(a))$ to the end point $(g(b), f(b))$ has slope

At a parameter value $t = c$, the velocity vector of the curve is $(g'(c), f'(c))$, and (since $g'(c) \neq 0$) the tangent line to the curve at that point has slope $f'(c)/g'(c)$. The Cauchy Mean Value Theorem says that somewhere along the curve, the tangent line is parallel to the chord — exactly the same geometric picture as the ordinary Mean Value Theorem, except now the curve need not be the graph of a function of $x$, since $g$ need not be monotonic in the trivial sense of being the identity. When $g(x) = x$, the curve $(g(t), f(t)) = (t, f(t))$ is just the graph of $f$, the chord slope becomes the ordinary difference quotient $\frac{f(b)-f(a)}{b-a}$, and we recover the familiar Mean Value Theorem. The Cauchy version is thus a re-parametrization of the same tangent-parallel-to-chord idea, applied to a curve traced by two functions simultaneously.

Example 1 — an explicit, asymmetric numerical case. Let $f(x) = x^3$ and $g(x) = x^2 + 1$ on the interval $[1, 3]$. Check the hypotheses. Both $f$ and $g$ are polynomials, hence continuous on $[1,3]$ and differentiable on $(1,3)$, with

On $(1,3)$, $x > 0$, so $g'(x) = 2x \neq 0$. All three hypotheses hold. Compute the left-hand side.

Solve for $c$. We need $c \in (1,3)$ with

Solving, $c = \frac{13}{4}\cdot\frac{2}{3} = \frac{13}{6} \approx 2.1667$. Verify $c \in (1,3)$: indeed $1 < 13/6 < 3$, since $13/6 \approx 2.17$. So $c = 13/6$ is the guaranteed point, and one checks directly:

This example is deliberately asymmetric (different-degree polynomials, an interval not centered anywhere special) so that no cancellation trick or symmetry hides an error — the algebra genuinely has to be tracked through.

Example 2 — a case where $g$ is not the identity and involves trigonometric functions. Let $f(x) = \sin x$ and $g(x) = \cos x$ on $[0, \pi/3]$. Check the hypotheses. $f$ and $g$ are continuous and infinitely differentiable everywhere, with

On $(0,\pi/3)$, $x \in (0, \pi/3) \subset (0,\pi)$, so $\sin x > 0$, hence $g'(x) = -\sin x \neq 0$. Hypotheses hold. Compute the left-hand side.

Solve for $c$. We need $c \in (0, \pi/3)$ with

So $\cot c = \sqrt3$, i.e. $\tan c = \dfrac{1}{\sqrt3}$, giving $c = \pi/6$. Verify $c \in (0,\pi/3)$: $\pi/6 \approx 0.524$ and $\pi/3 \approx 1.047$, and indeed $0 < \pi/6 < \pi/3$. Direct check:

Notice this recovers a classical fact: for $f = \sin$, $g=\cos$, the ratio $f'/g' = -\cot$, and the theorem is guaranteeing a mean-value point for the cotangent function — a nontrivial identity check beyond a routine polynomial example.