Hôpital's Rule

Differential calculus offers a way to calculate limits in the case of indeterminate forms of the type 0/0 and ∞/∞.
L'Hospital's Rule is named after a French nobleman, the Marquis de l'Hospital (1661–1704) a student of Johann Bernoulli, a marquis for whom the latter wrote the first textbook of analysis in the years 1691-1692. The portion of this textbook devoted to differential calculus was published in slightly altered form by l'Hopital under his own name in the "Analyse des infiniments petits" in 1696, the first text on differential calculus. Thus "l'Hopital's rule" is really due to Johann Bernoulli. L'Hôpital actually considered only the following special case in which two functions f, g are coninuos on [a,b], differentiable on (a,b) and both infinitesimal for x ⟶ a and showed that

where a is finite, and both f'(a) and g'(a) exist, finite, and are not zero. Indeed, using the definition of derivative

$$\frac{f^{\prime }(a)}{g^{\prime }(a)}=\frac{\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}}{\underset{h\to 0}{lim}\frac{g(a+h)-g(a)}{h}}=\underset{h\to 0}{lim}\frac{f(a+h)}{g(a+h)}$$

since f(a) = g(a) = 0 from the continuity of f,g at a. But this last condition is too restrictive. Indeed we have the following more general theorem.

Theorem 5.7.1 (L'Hôpital's Rule: 0/0 case). Let f and g be defined and differentiable on (x₀,b), such that g'(x) ≠ 0 ∀x ∈ (x₀,b). Suppose that

$$\underset{x\to x_0^{+}}{lim}f(x)=\underset{x\to x_0^{+}}{lim}g(x)=0$$

Then

$$\underset{x\to x_0^{+}}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=\ell \Rightarrow \underset{x\to x_0^{+}}{lim}\frac{f(x)}{g(x)}=\ell$$

Proof. Without loss of generality, it can be assumed that f(x₀) = g(x₀) = 0, because redefining the function at x₀ does not change the limits at x₀ of f, g, f', g' or their ratios (see example 5.7.2 for which changing the value of f(x) for x ∉ (a,b) won't change the limit). For x ≠ x₀, we may write

$$\frac{f(x)}{g(x)}=\frac{f(x)-f(x_0)}{g(x)-g(x_0)}$$

By Cauchy's Mean Value Theorem there is c between x and x₀ such that

$$\frac{f(x)}{g(x)}=\frac{f(x)-f(x_0)}{g(x)-g(x_0)}=\frac{f^{\prime }(c)}{g^{\prime }(c)}$$

Now as x ⟶ x₀, we must have also c ⟶ x₀ because c is between x and x₀. So if

$$\underset{x\to x_0}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=\ell$$

then

$$\underset{x\to x_0}{lim}\frac{f^{\prime }(c)}{g^{\prime }(c)}=\ell$$

Thefore

$$\underset{x\to x_0}{lim}\frac{f(x)}{g(x)}=\ell$$

as desired.  □

Note that there is nothing in this proof that requires ℓ to be finite. It works as well for the one-sided left-hand limits.

Example 5.7.2 Evaluate lim_{x ⟶ 0⁺} f(x), where f is

$$f=\{\begin{array}{ll}\frac{\cos 2x-\cos x}{x^2},& \text{if }x\in (0,\mathrm{\infty })\\ -2,& \text{if }x\notin (0,\mathrm{\infty }).\end{array}$$

Applying the de L’Hôpital’s Rule we have

Applying a second time de L’Hôpital’s Rule

Theorem 5.7.3 (L’Hôpital’s Rule for ∞/∞ Indeterminate Forms). Let f(x),g(x) two functions defined on a open interval I = (a,b) with a ∈ ℝ; Suppose that g is divergent to +∞ (−∞) for x ⟶ a⁻ and that both f,g are differentiable on I. Suppose also that g'(x) < 0 (g'(x) > 0) for all x ∈ I. Then if exists

$$\underset{x\to a^{+}}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=\ell$$

It holds also that

$$\underset{x\to a^{+}}{lim}\frac{f(x)}{g(x)}=\ell$$

Proof. We must show that for every decreasing sequence {a_n} of (a,b) such that lim_{n ⟶ ∞} x_n = a, results

$$\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_n)}{g(x_n)}=\ell$$

Suppose that g'(x) > 0 ∀x ∈ I: this implies that g(x) is an increasing function and consequenlty by Lemma 5.6.2 that {g(x_n)} is decreasing (if g'(x) < 0 it suffices to replace g and f by −g and −f and follow with the proof). By the Stolz–Cesàro theorem we have that the previous is verified if it results

$$\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}=\ell$$

From the differentiability of f,g ∈ I, we have that for every n ∈ ℕ there exits ξ_n ∈ (x_n, x_n+1) and by the Cauchy Mean Value theorem we have

$$\frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}=\frac{f^{\prime }({\xi }_n)}{g^{\prime }({\xi }_n)}$$

Since the sequence {ξ_n} converges as {x_n} to a⁺ we can deduce that being lim_{x ⟶ a⁺} f'(x)/g'(x) = ℓ, that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{f^{\prime }({\xi }_n)}{g^{\prime }({\xi }_n)}=\ell$$

which completes the proof.  □

Theorem 5.7.4 Let f(x) and g(x) two function defined on the same interval I = (a, +∞); Suppose f,g to be both infinitesimal for x ⟶ +∞ and differentiable on (a, +∞). Suppose also that g'(x) > 0 (g'(x) < 0) for all x ∈ I. Then if exists

$$\underset{x\to \mathrm{\infty }}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=\ell$$

Then it holds also that

$$\underset{x\to \mathrm{\infty }}{lim}\frac{f(x)}{g(x)}=\ell$$

Proof. Suppose withouth loss of generality that a > 0 and let x = 1/z, F(z) = f(1/z), G(z) = g(1/z). The functions F(z), G(z) are defined on the interval I' = (a,1/a) and infinitesimal for z ⟶ 0⁺. By the Chain Rule, the functions F,G are both differentiable ∀z ∈ (0,1/a), and we have

$$F^{\prime }(z)=\frac{-f^{\prime }(\frac{1}{z})}{z^2},G^{\prime }(z)=\frac{-g^{\prime }(\frac{1}{z})}{z^2},\frac{F^{\prime }(z)}{G^{\prime }(z)}=\frac{f^{\prime }(\frac{1}{z})}{g^{\prime }(\frac{1}{z})}$$

from which it results G'(z) ≠ 0 if z ∈ (0,1/a) and it exists:

$$\underset{z\to 0^{+}}{lim}\frac{F^{\prime }(z)}{G^{\prime }(z)}=\underset{x\to \mathrm{\infty }}{lim}\frac{f^{\prime }(z)}{g^{\prime }(z)}=\ell$$

We can thus apply Theorem 5.7.1 to functions F(z) and G(z) and deduce the existence of the limit

$$\underset{z\to 0^{+}}{lim}\frac{F(z)}{G(z)}=\underset{x\to \mathrm{\infty }}{lim}\frac{f(z)}{g(z)}=\ell$$

which proves the theorem.  □

Example 5.7.5. Evaluate the limit

$$\underset{x\to +\mathrm{\infty }}{lim}x\frac{\ln x+1}{\ln x-1}$$

We write

$$x\frac{\ln x+1}{\ln x-1}=\frac{\ln \frac{x+1}{x-1}}{1/x}$$

the right-hand side can also be written as

taking the derivative yields to

and lim_{x ⟶ +∞} 2x²/(x² − 1) = 2.  ■

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