Sign of f′ controls monotonicity
Here is the first dividend of the mean value theorem: f′ > 0 on an interval means f is strictly increasing there. This is the increasing-function test, and despite how obvious it feels, it genuinely needs the MVT — the derivative is a local quantity and increasing is a global statement, and only the MVT bridges them.
Claim: if f' > 0 on an interval I, then f is strictly increasing on I.
Proof. Take any x1 < x2 in I. Apply the MVT to f on [x1, x2]:
there is c in (x1, x2) with
f(x2) - f(x1) = f'(c) · (x2 - x1).
Now f'(c) > 0 (given) and x2 - x1 > 0, so the right side is > 0.
Hence f(x2) - f(x1) > 0, i.e. f(x2) > f(x1).
Since x1 < x2 were arbitrary, f is strictly increasing on I. ∎
Corollary (constancy): if f' = 0 on I, the same identity gives
f(x2) - f(x1) = 0 for all x1, x2, so f is constant on I.Cauchy's MVT and L'Hôpital
The Cauchy mean value theorem runs two functions at once: under the usual hypotheses there is a c with (f(b) − f(a)) g′(c) = (g(b) − g(a)) f′(c). Prove it by applying Rolle to the clever combination h(x) = (f(b) − f(a)) g(x) − (g(b) − g(a)) f(x), whose endpoints match. This two-function form is exactly the engine behind L'Hôpital's rule.
L'Hopital (0/0 form): suppose f, g differentiable near a, f(a) = g(a) = 0,
g'(x) ≠ 0 near a, and lim_{x->a} f'(x)/g'(x) = L. Then lim_{x->a} f(x)/g(x) = L.
Why, via Cauchy MVT. For x near a (say x > a), apply the Cauchy MVT on [a, x]:
there is c between a and x with
( f(x) - f(a) ) / ( g(x) - g(a) ) = f'(c) / g'(c).
But f(a) = g(a) = 0, so the left side is just f(x)/g(x). Hence
f(x) / g(x) = f'(c) / g'(c), with a < c < x.
As x -> a, the squeezed c -> a too, so f'(c)/g'(c) -> L.
Therefore f(x)/g(x) -> L. ∎
Worked instance: lim_{x->0} (sin x)/x. Here f=sin x, g=x, f(0)=g(0)=0,
f'/g' = (cos x)/1 -> cos 0 = 1. So the limit is 1 -- recovered honestly.