Cauchy's Mean Value Theorem

The mean value theorem survives on parametric curves, where two functions move together. A tangent parallel to the chord always exists.

Track the change of one quantity against another. Somewhere, the instantaneous ratio equals the whole interval's ratio. This is the bridge that proves l'Hôpital's rule.

For example, a 🚗 car climbing a mountain road gains distance and height together. At some moment it climbs at exactly the trip's overall height-to-distance ratio.

f′(ξ)g′(ξ)=f(b)−f(a)g(b)−g(a)\frac{f'(\xi)}{g'(\xi)} = \frac{f(b)-f(a)}{g(b)-g(a)}
ξ=149≈1.556\xi = \frac{14}{9} \approx 1.556

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Examples & Exercises

Prove the ordinary Mean Value Theorem is Cauchy's theorem with g(x)=xg(x) = x.

f(b)−f(a)g(b)−g(a)=f′(ξ)g′(ξ)→g(x)=xf′(ξ)=f(b)−f(a)b−a\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(\xi)}{g'(\xi)} \xrightarrow{g(x)=x} f'(\xi) = \frac{f(b)-f(a)}{b-a}
ξ=14/9≈1.56\xi = 14/9 \approx 1.56: the tangent ratio 2/(3ξ)=3/72/(3\xi) = 3/7 matches the chord slope.
Show proof

  1. Cauchy's theorem: if f,gf, g are continuous on [a,b][a, b], differentiable on (a,b)(a, b) with g′≠0g' \neq 0, then some ξ∈(a,b)\xi \in (a, b) satisfies f(b)−f(a)g(b)−g(a)=f′(ξ)g′(ξ)\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(\xi)}{g'(\xi)}.

  2. Substituting g(x)=xg(x) = x gives g(b)−g(a)=b−ag(b) - g(a) = b - a and g′(ξ)=1g'(\xi) = 1, so the formula becomes f(b)−f(a)b−a=f′(ξ)1=f′(ξ)\frac{f(b)-f(a)}{b-a} = \frac{f'(\xi)}{1} = f'(\xi).

  3. That is exactly the ordinary Mean Value Theorem — the Cauchy form with the special choice g(x)=xg(x) = x.

Prove lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x} = 1 with Cauchy's Mean Value Theorem.

sin⁡x−sin⁡0x−0=cos⁡ξx1,0<ξx<x  ⇒  cos⁡ξx→1\frac{\sin x - \sin 0}{x - 0} = \frac{\cos \xi_x}{1}, \quad 0 < \xi_x < x \;\Rightarrow\; \cos \xi_x \to 1
ξ=14/9≈1.56\xi = 14/9 \approx 1.56 — the theorem point where the ratio equals the chord slope 37\frac37.
Show proof

  1. For x>0x > 0, sin⁡xx=sin⁡x−sin⁡0x−0\frac{\sin x}{x} = \frac{\sin x - \sin 0}{x - 0}.

    Both f(t)=sin⁡tf(t) = \sin t and g(t)=tg(t) = t are continuous on [0,x][0, x] and differentiable on (0,x)(0, x), so Cauchy's hypotheses hold.

    Applying Cauchy's theorem to them on [0,x][0, x] (with g′=1≠0g' = 1 \neq 0), some ξx∈(0,x)\xi_x \in (0, x) gives sin⁡x−sin⁡0x−0=cos⁡ξx1=cos⁡ξx\frac{\sin x - \sin 0}{x - 0} = \frac{\cos \xi_x}{1} = \cos \xi_x.

  2. Since 0<ξx<x0 < \xi_x < x, squeezing gives ξx→0\xi_x \to 0 as x→0+x \to 0^+. Cosine is continuous (differentiable implies continuous), so cos⁡ξx→cos⁡0=1\cos \xi_x \to \cos 0 = 1.

  3. For x<0x < 0, the same theorem on [x,0][x, 0] yields ξx∈(x,0)\xi_x \in (x, 0) and likewise cos⁡ξx→1\cos \xi_x \to 1. Hence lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x} = 1.

Prove lim⁡x→0ex−1−xx2=12\lim_{x \to 0}\frac{e^x - 1 - x}{x^2} = \frac12 by applying Cauchy's theorem twice. (State the 0/00/0 hypotheses.)

ex−1−xx2=eξ−12ξ=eξ22→x→0+12\frac{e^x-1-x}{x^2} = \frac{e^{\xi}-1}{2\xi} = \frac{e^{\xi_2}}{2} \xrightarrow{x \to 0^+} \frac12
ξ=14/9≈1.56\xi = 14/9 \approx 1.56 — note there is only one theorem point, because the ratio 2/(3t)2/(3t) is strictly decreasing.
Show proof

  1. Put f(x)=ex−1−xf(x) = e^x - 1 - x and g(x)=x2g(x) = x^2: then f(0)=g(0)=0f(0) = g(0) = 0, both are continuous on [0,x][0, x] (for x>0x > 0), differentiable on (0,x)(0, x), and g′(t)=2t≠0g'(t) = 2t \neq 0.

  2. Cauchy's theorem gives f(x)g(x)=f′(ξ)g′(ξ)=eξ−12ξ\frac{f(x)}{g(x)} = \frac{f'(\xi)}{g'(\xi)} = \frac{e^{\xi}-1}{2\xi} for some ξ∈(0,x)\xi \in (0, x). Applying it again to f1(t)=et−1f_1(t) = e^t - 1, g1(t)=2tg_1(t) = 2t on [0,ξ][0, \xi] (with f1(0)=g1(0)=0f_1(0) = g_1(0) = 0 and g1′=2≠0g_1' = 2 \neq 0) yields eξ−12ξ=eξ22\frac{e^{\xi}-1}{2\xi} = \frac{e^{\xi_2}}{2} for some ξ2∈(0,ξ)\xi_2 \in (0, \xi).

  3. Since 0<ξ2<ξ<x0 < \xi_2 < \xi < x, x→0+x \to 0^+ forces ξ2→0\xi_2 \to 0, and continuity of ete^t gives eξ2→1e^{\xi_2} \to 1.

    So the ratio tends to 12\frac12.

    The same argument on [x,0][x, 0] covers x<0x < 0, hence lim⁡x→0ex−1−xx2=12\lim_{x \to 0}\frac{e^x-1-x}{x^2} = \frac12.

10. Differentiation