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Math Analysis 1

The theorems of Fermat, Rolle, Lagrange and Cauchy, and l'Hôpital's rule

Stationary points, Rolle's theorem, the mean value theorem and its consequences for monotonicity, Cauchy's theorem, l'Hôpital's rule with examples and when not to use it.

The theorems of Fermat, Rolle, Lagrange and Cauchy link the derivative to the global behaviour of a function: monotonicity, maxima and minima, estimates. L’Hôpital’s rule, which follows from them, is the most used tool for indeterminate forms.

Fermat: stationary points

If f has a local maximum or minimum at an interior point x_0 and is differentiable there, then f'(x_0) = 0. The converse is false: x^3 has zero derivative at 0 but neither a maximum nor a minimum there. That is why the candidates for extrema of a function on an interval are three kinds of points: those with f' = 0, those where f is not differentiable, and the endpoints of the interval.

Rolle

If f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), there is c \in (a, b) with f'(c) = 0. Proof: by the extreme value theorem f has a maximum and a minimum on [a, b]; if both are at the endpoints the function is constant, otherwise one of them is interior and by Fermat the derivative vanishes there.

Lagrange (mean value theorem)

If f is continuous on [a, b] and differentiable on (a, b), there is c \in (a, b) with

    \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]

Somewhere the tangent is parallel to the chord. In terms of speed: if you cover 90 km in one hour, at least at one moment the speedometer showed exactly 90 km/h. The proof applies Rolle to f(x) minus the line of the chord.

Consequences used all the time:

  • if f' = 0 on an interval, f is constant there (which is why antiderivatives differ by a constant);
  • if f' > 0 on an interval, f is strictly increasing; if f' < 0, decreasing;
  • estimates: |\sin a - \sin b| \le |a - b|, because |\cos c| \le 1.

Cauchy

It extends Lagrange to two functions: with the same hypotheses on f and g, and g' \ne 0, there is c with \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}. It is the step needed to prove l’Hôpital’s rule.

L’Hôpital’s rule

If f(x) and g(x) both tend to 0 (or both to \pm\infty) as x \to x_0, are differentiable near x_0 with g' \ne 0, and the limit of \frac{f'}{g'} exists, then

    \[ \lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f'(x)}{g'(x)}. \]

It also holds for x_0 = \pm\infty.

  • \displaystyle\lim_{x\to0}\frac{x - \sin x}{x^3} = \lim\frac{1 - \cos x}{3x^2} = \lim\frac{\sin x}{6x} = \frac16 (applied three times);
  • \displaystyle\lim_{x\to+\infty}\frac{\ln x}{\sqrt x} = \lim\frac{1/x}{1/(2\sqrt x)} = \lim\frac{2}{\sqrt x} = 0;
  • \displaystyle\lim_{x\to0^+} x\ln x: it is 0\cdot\infty, rewrite it as \frac{\ln x}{1/x} and get \lim\frac{1/x}{-1/x^2} = \lim(-x) = 0.

When not to use it

  • When there is no indeterminate form: \frac{x+1}{x+2} as x\to0 is \frac12; differentiating would give 1, which is wrong.
  • When the limit of f'/g' does not exist nothing can be concluded: \frac{x + \sin x}{x} as x\to\infty tends to 1, but \frac{1 + \cos x}{1} has no limit.
  • To prove the standard limits: you would need to know the derivatives already, and they are derived from those very limits.
  • When the derivatives get messy: equivalences or Taylor expansions often reach the result in one line.
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