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L’Hôpital’s rule

Summary

This lesson proves L’Hôpital’s rule: if ff and gg tend to 00, or if gg tends to infinity, and the quotient of their derivatives has a limit, then the quotient fg\frac{f}{g} has the same limit. The proof applies Cauchy’s mean value theorem, from the preceding lesson, on an interval with one endpoint fixed and the other variable; the case in which the denominator tends to infinity requires a finer argument, which asks nothing of the numerator. The rule holds on one side of a point, at the point and at infinity, and can be applied several times in a row: this is how one computes the limit of f(x)−f(a)(x−a)n\frac{f(x) - f(a)}{(x - a)^n} when the first derivatives of ff vanish at aa. The forms 0⋅∞0 \cdot \infty, ∞−∞\infty - \infty, 1∞1^\infty, 000^0 and ∞0\infty^0 are reduced to a quotient. The lesson also shows when the rule does not apply or is inconclusive, why it cannot be used for the limits with which the derivatives were computed, and how it combines with equivalent infinitesimals.

Learning objectives

  1. Prove L’Hôpital’s rule for the forms 00\frac{0}{0} and ∞∞\frac{\infty}{\infty}, on one side of a point, at the point and at infinity, from Cauchy’s mean value theorem.
  2. Compute limits of the indeterminate forms by reducing them to a quotient, and apply the rule several times, checking its hypotheses at each application.
  3. Recognize when the rule does not apply or is inconclusive: limits that are not indeterminate, quotients of derivatives without a limit, derivatives of the denominator that vanish near the point, and the limits that the computation of the derivatives presupposes.
  4. Combine the rule with equivalent infinitesimals and with the algebra of limits, and choose in each case the shortest route.

The form zero over zero

The corollary on the limit of a quotient, from the lesson on the algebra of limits, requires the denominator to have a nonzero limit. If the numerator and the denominator both tend to 00, the quotient can tend to any number, or to infinity, or have no limit: 3xx\frac{3x}{x} tends to 33, x2x\frac{x^2}{x} tends to 00, and xsin⁡1xx\frac{x \sin \frac{1}{x}}{x} has no limit at 00. That is why 00\frac{0}{0} is called an indeterminate form: the limit of each term does not determine that of the quotient. The lesson on equivalent infinitesimals solved many of these limits by replacing each factor with an equivalent one from its table; but a difference such as x−sin⁡xx - \sin x is not a product of factors from the table, and its terms cannot be replaced.

The derivative suggests another route. If f(a)=g(a)=0f(a) = g(a) = 0, if ff and gg are differentiable at aa and if g′(a)≠0g'(a) \neq 0, then dividing the numerator and the denominator by x−ax - a turns the quotient into a quotient of difference quotients, and the corollary on the limit of a quotient gives:

f(x)g(x)=f(x)−f(a)x−ag(x)−g(a)x−a→f′(a)g′(a)\frac{f(x)}{g(x)} = \frac{\frac{f(x) - f(a)}{x - a}}{\frac{g(x) - g(a)}{x - a}} \to \frac{f'(a)}{g'(a)}

But often g′(a)g'(a) is also zero, or ff and gg are not defined at aa. L’Hôpital’s rule replaces the values of the derivatives at aa with the limit of their quotient near aa. The tool is Cauchy’s mean value theorem, from the lesson on the theorems of Fermat, Rolle, Lagrange and Cauchy: if ff and gg are continuous on [a,b]\icc{a}{b}, with a<ba < b, and differentiable at every point of (a,b)\ioo{a}{b}, and if g′(x)≠0g'(x) \neq 0 for every xx in (a,b)\ioo{a}{b}, then g(b)≠g(a)g(b) \neq g(a) and there exists a cc in (a,b)\ioo{a}{b} such that f(b)−f(a)g(b)−g(a)=f′(c)g′(c)\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}. Unlike two applications of Lagrange’s theorem, which would give two different points, Cauchy’s theorem gives one and the same point for the numerator and for the denominator.

TheoremL’Hôpital’s rule for the form zero over zero

Let a<ba < b, and let ff and gg be functions differentiable at every point of (a,b)\ioo{a}{b}, with g′(x)≠0g'(x) \neq 0 for every xx in (a,b)\ioo{a}{b}. Suppose that f(x)f(x) and g(x)g(x) tend to 00 as x→a+x \to a^+, and that

lim⁡x→a+f′(x)g′(x)=L\lim_{x \to a^+} \frac{f'(x)}{g'(x)} = L

where LL is a real number, ∞\infty or −∞-\infty. Then g(x)≠0g(x) \neq 0 for every xx in (a,b)\ioo{a}{b}, and

lim⁡x→a+f(x)g(x)=L\lim_{x \to a^+} \frac{f(x)}{g(x)} = L

One extends ff and gg with the value 00 at aa, and Cauchy’s theorem is applied on [a,x]\icc{a}{x}: the quotient f(x)g(x)\frac{f(x)}{g(x)} is a quotient of increments, equal to that of the derivatives at a point between aa and xx. The figure shows, with f(x)=sin⁡xf(x) = \sin x, g(x)=ex−1g(x) = e^x - 1 and a=0a = 0, the points (g(t),f(t))\left(g(t), f(t)\right): the quotient f(x)g(x)\frac{f(x)}{g(x)} is the slope of the chord joining the origin to the point corresponding to xx, and the quotient of the derivatives at cc is the slope of the parallel tangent.

Proof

  1. F(a)=G(a)=0,F′(t)=f′(t),G′(t)=g′(t)≠0,a<t<xF(a) = G(a) = 0, \qquad F'(t) = f'(t), \quad G'(t) = g'(t) \neq 0, \qquad a < t < x

    Let FF and GG be the functions equal to ff and gg on (a,b)\ioo{a}{b} and to 00 at aa. Let xx be in (a,b)\ioo{a}{b}. Both are continuous on [a,x]\icc{a}{x}: at aa, because their right-hand limits are 00; at the other points, by the theorem that differentiable implies continuous, from the lesson on the derivative as a limit. At every point of (a,x)\ioo{a}{x} they are differentiable, with the derivatives of ff and of gg, by the lemma on the local character of the derivative, from the lesson on differentiation rules.

  2. f(x)g(x)=F(x)−F(a)G(x)−G(a)=f′(c)g′(c),a<c<x\frac{f(x)}{g(x)} = \frac{F(x) - F(a)}{G(x) - G(a)} = \dato{c}{\frac{f'(c)}{g'(c)}}, \qquad a < c < x

    By Cauchy’s mean value theorem on [a,x]\icc{a}{x}, G(x)≠G(a)G(x) \neq G(a), that is, g(x)≠0g(x) \neq 0, and there exists a cc between aa and xx such that the quotient of increments equals that of the derivatives at cc.

  3. a<x<a+δ→∣f(x)g(x)−L∣=∣f′(c)g′(c)−L∣<εa < x < a + \delta \rightarrow \abs{\frac{f(x)}{g(x)} - L} = \abs{\frac{f'(c)}{g'(c)} - L} < \resaltar{\varepsilon}

    Let LL be real and let ε>0\varepsilon > 0. By the definition of the right-hand limit, from the lesson on one-sided limits, there exists a δ>0\delta > 0, which can be taken less than or equal to b−ab - a, such that ∣f′(t)g′(t)−L∣<ε\abs{\frac{f'(t)}{g'(t)} - L} < \varepsilon if a<t<a+δa < t < a + \delta. If a<x<a+δa < x < a + \delta, the point cc from the second move satisfies a<c<x<a+δa < c < x < a + \delta. If L=∞L = \infty, the same argument with a challenge NN gives f(x)g(x)>N\frac{f(x)}{g(x)} > N; if L=−∞L = -\infty, it gives f(x)g(x)<N\frac{f(x)}{g(x)} < N.

It is not necessary to know how cc depends on xx, nor that it is unique: it suffices that it lies between aa and xx. The hypothesis g′≠0g' \neq 0 is required on the whole interval (a,b)\ioo{a}{b}, because Cauchy’s theorem uses it on each [a,x]\icc{a}{x}; since the limit depends only on nearby values, it suffices to find an interval (a,b)\ioo{a}{b} where it holds, however small. It is not enough, on the other hand, that the derivative of the denominator vanish only at isolated points that accumulate at the point, or that are unbounded if the variable tends to infinity, as solved problem 5 shows. If ff and gg are defined on a larger set, the theorem is applied to their restrictions to (a,b)\ioo{a}{b}, which have the same right-hand limit at aa.

The same holds on the left, with an interval (b,a)\ioo{b}{a}, b<ab < a, and x→a−x \to a^-: it suffices to apply the theorem to f(−t)f(-t) and g(−t)g(-t) near −a-a on the right, by the translation and reflection proposition, from the lesson on one-sided limits; the chain rule changes the sign of both derivatives and does not alter their quotient. If the hypotheses hold on both sides of aa, on (a−r,a)\ioo{a - r}{a} and on (a,a+r)\ioo{a}{a + r}, the theorem on the limit and one-sided limits, from that lesson, gives the limit as x→ax \to a (if LL is infinite, with the same argument of the smaller of the two radii). The function may be defined at aa or not; its value there plays no role.

ExampleThree applications in a row

Consider lim⁡x→0x−sin⁡xx3\lim_{x \to 0} \frac{x - \sin x}{x^3}. The numerator and the denominator tend to 00, and the derivative of the denominator, 3x23x^2, does not vanish away from 00. The quotient of the derivatives is 1−cos⁡x3x2\frac{1 - \cos x}{3x^2}, again of the form 00\frac{0}{0}, with a denominator whose derivative is 6x6x, which does not vanish away from 00 either; and the new quotient, sin⁡x6x\frac{\sin x}{6x}, tends to 16\frac{1}{6} by the fundamental trigonometric limit:

lim⁡x→0x−sin⁡xx3=lim⁡x→01−cos⁡x3x2=lim⁡x→0sin⁡x6x=16\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{1 - \cos x}{3x^2} = \lim_{x \to 0} \frac{\sin x}{6x} = \frac{1}{6}

The sequence of equalities is read from right to left: the last limit exists, and the rule, applied twice, transmits its value backward. Writing the first equals sign before knowing that the second limit exists is only a way of recording the computation, which is justified once it is finished. The second limit also follows from the table of equivalents, with 1−cos⁡x∼x221 - \cos x \sim \frac{x^2}{2}; the first does not.