Cauchy's MVT: For f, g continuous on [a,b], differentiable on (a,b), g'(x) != 0 on (a,b), there exists c in (a,b) with f'$\frac{c}{g}$'(c) = [f(b)-f(a)]/[g(b)-g(a)].

Connection to L'Hopital's Rule: For the 0/0 form (f(a) = g(a) = 0): f$\frac{x}{g}$(x) = [f(x)-f(a)]/[g(x)-g(a)] = f'(c(x))/g'(c(x)) where c(x) is between a and x. As x -> a, c(x) -> a. If lim f'$\frac{x}{g}$'(x) exists as x -> a, it equals lim f$\frac{x}{g}$(x).

Why g'(x) != 0 matters: Without this, g could have the same value at two points, making the denominator zero. It also ensures the ratio f'$\frac{c}{g}$'(c) is defined.

Parametric interpretation: If x = g(t), y = f(t) traces a curve, Cauchy's MVT says the slope of the tangent dy/dx at some point equals the slope of the chord joining endpoints.

Verification example: f(x) = sin x, g(x) = cos x on [0, pi/2]. LHS: f'$\frac{c}{g}$'(c) = cos$\frac{c}{(-sin(c)}$) = -cot(c). RHS: (sin$\frac{pi}{2}$-sin(0))/(cos$\frac{pi}{2}$-cos(0)) = $\frac{1}{-1}$ = -1. So cot(c) = 1, c = pi/4 in (0, pi/2).