L'Hôpital's rule

Guillaume de l'Hôpital, after whom this rule is named.

In calculus, l'Hôpital's rule pronounced: [lopiˈtal] (also called Bernoulli's rule) uses derivatives to help evaluate limits involving indeterminate forms. Application (or repeated application) of the rule often converts an indeterminate form to a determinate form, allowing easy evaluation of the limit. The rule is named after the 17th-century French mathematician Guillaume de l'Hôpital, who published the rule in his book Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes (literal translation: Analysis of the Infinitely Small to Understand Curved Lines) (1696), the first textbook on differential calculus.^[1][2] However, it is believed that the rule was discovered by the Swiss mathematician Johann Bernoulli.^[3]

The Stolz-Cesàro theorem is a similar result involving limits of sequences, but it uses finite difference operators rather than derivatives.

In its simplest form, l'Hôpital's rule states that for functions f and g:

If   $\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0$ or $\pm\infty$   and   $\lim_{x\to c}\frac{f'(x)}{g'(x)}$   exists,

then   $\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$

The differentiation of the numerator and denominator often simplifies the quotient and/or converts it to a determinate form, allowing the limit to be evaluated more easily.

General form

The general form of l'Hôpital's rule covers many more cases. Suppose that c and L are extended real numbers (i.e., real numbers, positive infinity, or negative infinity). Suppose that either

$\lim_{x\to c}{f(x)} = \lim_{x\to c}g(x) = 0$

or

$\lim_{x\to c}{f(x)} = \pm\lim_{x\to c}{g(x)} = \pm\infty.$

And suppose that

$\lim_{x\to c}{\frac{f'(x)}{g'(x)}} = L.$

Then

$\lim_{x\to c}{\frac{f(x)}{g(x)}}=L.$

The limits may also be one-sided limits.

Requirement that limit exist

The requirement that the limit

$\lim_{x\to c}\frac{f'(x)}{g'(x)}$

exists is essential. Differentiation of indeterminate forms can sometimes lead to limits that do not exist. If this happens, then l'Hôpital's rule does not apply. For example, if f(x) = x + sin(x) and g(x) = x, then

$\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=\lim_{x\to\infty}\frac{1+\cos x}{1},$

which limit does not exist. However, working with the original functions

$\lim_{x\to\infty}\frac{f(x)}{g(x)} = \lim_{x\to\infty}\left(1+\frac{\sin x}{x}\right) = 1$.

Examples

• Here is an example involving the sinc function and the indeterminate form ⁰⁄₀:

$\begin{align} \lim_{x \to 0} \operatorname{sinc}(x) & = \lim_{x \to 0} \frac{\sin\pi x}{\pi x} \\ & = \lim_{y \to 0} \frac{\sin y}{y} \\ & = \lim_{y \to 0} \frac{\cos y}{1} \\ & = 1. \end{align}$

Alternatively, just observe that the limit is the definition of the derivative of the sine function at zero.

• This is a more elaborate example involving ⁰⁄₀. Applying l'Hôpital's rule a single time still results in an indeterminate form. In this case, the limit may be evaluated by applying the rule three times:

$\begin{align} \lim_{x\to 0}{\frac{2\sin x-\sin 2x}{x-\sin x}} & =\lim_{x\to 0}{\frac{2\cos x -2\cos 2x}{1-\cos x}} \\ & = \lim_{x\to 0}{\frac{-2\sin x +4\sin 2x}{\sin x}} \\ & = \lim_{x\to 0}{\frac{-2\cos x +8\cos 2x}{\cos x}} \\ & ={\frac{-2 +8}{1}} \\ & =6. \end{align}$

• This example involves ⁰⁄₀. Suppose that b > 0. Then

$\lim_{x\to 0} {\frac{b^x - 1}{x}}=\lim_{x \to 0}{\frac{b^x \ln b}{1}}=\ln b \lim_{x \to 0}{b^x}=\ln b.$

• Here is another example involving ⁰⁄₀:

$\lim_{x\to 0}{\frac{e^x-1-x}{x^2}} =\lim_{x\to 0}{\frac{e^x-1}{2x}} =\lim_{x\to 0}{\frac{e^x}{2}}={\frac{1}{2}}.$

• This example involves ^∞⁄_∞. Assume n is a positive integer. Then

$\lim_{x\to\infty} x^n e^{-x} =\lim_{x\to\infty}{\frac{x^n}{e^x}} =\lim_{x\to\infty}{\frac{nx^{n-1}}{e^x}} =n\lim_{x\to\infty}{\frac{x^{n-1}}{e^x}}.$

Repeatedly apply l'Hôpital's rule until the exponent is zero to conclude that the limit is zero.

• Here is another example involving ^∞⁄_∞:

$\lim_{x\to 0^+} x \ln x =\lim_{x\to 0^+}{\frac{\ln x}{1/x}} =\lim_{x\to 0^+}{\frac{1/x}{-1/x^2}} =\lim_{x\to 0^+} -x = 0.$

• Here is an example involving the impulse response of a raised-cosine filter and ⁰⁄₀:

$\begin{align} \lim_{t\to 1/2} \operatorname{sinc}(t) \frac{\cos \pi t}{1 - (2t)^2} & = \operatorname{sinc}(1/2) \lim_{t\to 1/2} \frac{\cos \pi t}{1 - (2 t)^2} \\ & = \frac{2}{\pi} \lim_{t\to 1/2} \frac{-\pi \sin \pi t}{-8 t} \\ & = \frac{2}{\pi} \cdot \frac{\pi}{4} \\ & = \frac{1}{2}. \end{align}$

• One can also use l'Hôpital's rule to prove the following theorem. If f ′′ is continuous at x, then

$\begin{align} \lim_{h \to 0} \frac{f(x + h) + f(x - h) - 2f(x)}{h^2} & = \lim_{h \to 0} \frac{f'(x + h) - f'(x - h)}{2h} \\ & = f''(x). \end{align}$

• Sometimes L'Hôpital's rule is invoked in a tricky way: suppose f(x) + f ′(x) converges as x → ∞. It follows:
•  :$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} {e^x f(x) \over e^x} = \lim_{x \to \infty} {e^x (f(x) + f'(x)) \over e^x} = \lim_{x \to \infty} (f(x) + f'(x))$

and so   $\lim_{x \to \infty} f(x)$   exists and   $\lim_{x \to \infty} f'(x) = 0.$

Note that the above proof assume that e^xf(x) converges to positive or negative infinity as x → ∞. Hence, we can then apply L'Hôpital's rule. The results however still hold even though the proof is not entirely complete.

Other indeterminate forms

Other indeterminate forms, such as 1^∞, 0⁰, ∞⁰, 0 × ∞, and ∞ − ∞, can sometimes be evaluated using l'Hôpital's rule. For example, to evaluate a limit involving ∞ − ∞, convert the difference of two functions to a quotient:

$\begin{align} \lim_{x \to 1} \left( \frac{x}{x-1} - \frac{1}{\ln x} \right) & = \lim_{x \to 1} \frac{x \ln x - x + 1}{(x-1) \ln x} \quad (1) \\ & = \lim_{x \to 1} \frac{\ln x}{\frac{x-1}{x} + \ln x} \quad (2) \\ & = \lim_{x \to 1} \frac{x \ln x}{x - 1 + x \ln x} \quad (3) \\ & = \lim_{x \to 1} \frac{1 + \ln x}{1 + 1 + \ln x} \quad (4) \\ & = \lim_{x \to 1} \frac{1 + \ln x}{2 + \ln x} \\ & = \frac{1}{2}, \end{align}$

where l'Hôpital's rule was applied in going from (1) to (2) and then again in going from (3) to (4).

l'Hôpital's rule can be used on indeterminate forms involving exponents by using logarithms to "move the exponent down". Here is an example involving the indeterminate form 0⁰:

$\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{\ln x^x} = \lim_{x \to 0^+} e^{x \ln x} = e^{\lim_{x \to 0^+} (x \ln x)}.$

It is valid to move the limit inside the exponential function because the exponential function is continuous. Now the exponent x has been "moved down". The limit lim_x→0⁺ (x ln x) is of the indeterminate form 0 × (-∞), but as shown in an example above, l'Hôpital's rule may be used to determine that

$\lim_{x \to 0^+} x \ln x = 0.$

Thus

$\lim_{x \to 0^+} x^x = e^0 = 1.$

Other methods of evaluating limits

Although l'Hôpital's rule is a powerful way of evaluating otherwise hard-to-evaluate limits, it is not always the easiest way. Consider

$\lim_{|x| \to \infty} x \sin \frac{1}{x}.$

This limit may be evaluated using l'Hôpital's rule:

$\begin{align} \lim_{|x| \to \infty} x \sin \frac{1}{x} & = \lim_{|x| \to \infty} \frac{\sin \frac{1}{x}}{1/x} \\ & = \lim_{|x| \to \infty} \frac{-x^{-2}\cos\frac{1}{x}}{-x^{-2}} \\ & = \lim_{|x| \to \infty} \cos\frac{1}{x} \\ & = \cos{\left(\lim_{|x| \to \infty} \frac{1}{x} \right)} \\ & = 1. \end{align}$

It is valid to move the limit inside the cosine function because the cosine function is continuous.

Another way to evaluate this limit is to use a substitution. y = 1/x. As |x| approaches infinity, y approaches zero. So,

$\lim_{|x| \to \infty} x \sin \frac{1}{x} = \lim_{y \to 0} \frac{\sin y}{y} = 1.$

The final limit may be evaluated using l'Hôpital's rule or by noting that it is the definition of the derivative of the sine function at zero.

Still another way to evaluate this limit is to use a Taylor series expansion:

$\begin{align} \lim_{|x| \to \infty} x \sin \frac{1}{x} & = \lim_{|x| \to \infty} x \left( \frac{1}{x} - \frac{1}{3!\, x^3} + \frac{1}{5!\, x^5} - \cdots \right) \\ & = \lim_{|x| \to \infty} 1 - \frac{1}{3!\, x^2} + \frac{1}{5!\, x^4} - \cdots \\ & = 1 + \lim_{|x| \to \infty} \frac{1}{x}\left(-\frac{1}{3!\, x} + \frac{1}{5!\, x^3} - \cdots \right). \end{align}$

For |x| ≥ 1, the expression in parentheses is bounded, so the limit in the last line is zero.

Logical circularity

In some cases it may constitute circular reasoning to use l'Hôpital's rule to evaluate a limit. Consider

$\lim_{h\to 0} \frac{(x+h)^n-x^n}{h}.$

If the purpose of evaluating this limit is to prove that if f(x) = xⁿ, then

$f'(x)=nx^{n-1},\,$

and one uses l'Hôpital's rule and this same fact to evaluate the limit, then the argument uses the conclusion as an assumption (i.e., begging the question) and is therefore fallacious (even though the conclusion is true).

Case where f and g are continuously differentiable

The proof of l'Hôpital's rule is simple in the case where f and g are continuously differentiable at the point c. It is not a proof of the general l'Hôpital's rule because it requires stronger hypotheses than does l'Hôpital's rule.

Suppose that f and g are continuously differentiable at c, that f(c) = g(c) = 0, and that g′(c) ≠ 0. Then

$\lim_{x\to c}\frac{f(x)}{g(x)}= \lim_{x\to c}\frac{f(x)-f(c)}{g(x)-g(c)} = \lim_{x\to c}\frac{\left(\frac{f(x)-f(c)}{x-c}\right)}{\left(\frac{g(x)-g(c)}{x-c}\right)} =\frac{\lim_{x\to c}\left(\frac{f(x)-f(c)}{x-c}\right)}{\lim_{x\to c}\left(\frac{g(x)-g(c)}{x-c}\right)}= \frac{f'(c)}{g'(c)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$

(remember that f(c) = g(c) = 0). This follows from the limit rules for quotients and the definition of the derivative. The last equality uses continuity of the derivatives at c.

This suggests the general case of l'Hôpital's rule, which does not require the functions f and g to be continuously differentiable at the point c and is proven below.

Geometric interpretation

Consider the curve in the plane whose x-coordinate is given by g(t) and whose y-coordinate is given by f(t), i.e.

$t\mapsto [g(t),f(t)]. \,$

Suppose f(c) = g(c) = 0. The limit of the ratio f(t)/g(t) as t → c is the slope of tangent to the curve at the point [0, 0]. The tangent to the curve at the point t is given by [g′(t), f ′(t)]. L'Hôpital's rule then states that the slope of the tangent at 0 is the limit of the slopes of tangents at the points approaching zero.

Proof of l'Hôpital's rule

A standard proof of l'Hôpital's rule uses Cauchy's mean value theorem. l'Hôpital's rule has many variations depending on whether c and L are finite or infinite, whether f and g converge to zero or infinity, and whether the limits are one-sided or two-sided. All the variations follow from the two main variations below without, for the most part, requiring any new reasoning.^[4]

Zero over zero

Suppose that c and L are finite and f and g converge to zero.

First, define (or redefine) f(c) = 0 and g(c) = 0. This makes f and g continuous at c, but does not change the limit (since, by definition, the limit does not depend on the value at the point c). Since $\lim_{x\to c}f'(x)/g'(x)$ exists, there is an interval (c − δ, c + δ) such that for all x in the interval, with the possible exception of x = c, both f ′(x) and g′(x) exist and g′(x) is not zero.

If x is in the interval (c, c + δ), then the mean value theorem and Cauchy's mean value theorem both apply to the interval [c, x], and a similar statement holds for x in the interval (c − δ, c). The mean value theorem implies that g(x) is not zero, since otherwise there would be some y in the interval (c, x) with g′(y) = 0. Cauchy's mean value theorem now implies that there is a point ξ_x in (c, x) such that

$\frac{f(x)}{g(x)} = \frac{f'(\xi_x)}{g'(\xi_x)}.$

If x approaches c, then ξ_x approaches c (by the squeeze theorem). Since $\lim_{x\to c}f'(x)/g'(x)$ exists, it follows that

$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(\xi_x)}{g'(\xi_x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$

Infinity over infinity

Suppose that L is finite, c is positive infinity, and f and g converge to positive infinity.

For every ε > 0, there is an m such that

$\left|\frac{f'(x)}{g'(x)} - L\right| < \varepsilon \quad \text{for } x\geq m.$

The mean value theorem implies that if x > m, then g(x) ≠ g(m), since otherwise there would be a y in the interval (m, x) with g′(y) = 0. Cauchy's mean value theorem applied to the interval [m, x] now implies that

$\left|\frac{f(x)-f(m)}{g(x)-g(m)} - L\right| < \varepsilon \quad \text{for } x>m.$

Since f converges to positive infinity, if x is large enough, then f(x) ≠ f(m). Write

$\frac{f(x)}{g(x)} = \frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)}.$

Now,

$\begin{align} & \left|\frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - \frac{f(x)-f(m)}{g(x)-g(m)}\right| \\ & \quad \leq \left|\frac{f(x)-f(m)}{g(x)-g(m)}\right| \left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right| \\ & \quad < (|L|+\varepsilon)\left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right|. \end{align}$

For x sufficiently large, this is less than ε and therefore

$\left|\frac{f(x)}{g(x)} - L\right| < 2\varepsilon.$*

• Note: To arrive at the above inequality add :$\begin{align} & \left|\frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - \frac{f(x)-f(m)}{g(x)-g(m)}\right| \\\end{align}$

and

$\left|\frac{f(x)-f(m)}{g(x)-g(m)} - L\right|$

Then apply the triangle inequality.

• Note: Steps are missing.

See also

• l'Hôpital controversy

Notes

1. ^ O'Connor, John J.; Robertson, Edmund F. "De L'Hopital biography". The MacTutor History of Mathematics archive. Scotland: School of Mathematics and Statistics, University of St Andrews. http://www-history.mcs.st-andrews.ac.uk/Biographies/De_L'Hopital.html. Retrieved 21 December 2008. 
2. ^ l’Hospital, Analyse des infiniment petits… , pages 145-146: “Proposition I. Problême. Soit une ligne courbe AMD (AP = x, PM = y, AB = a [see Figure 130] ) telle que la valeur de l’appliquée y soit exprimée par une fraction, dont le numérateur & le dénominateur deviennent chacun zero lorsque x = a, c’est à dire lorsque le point P tombe sur le point donné B. On demande quelle doit être alors la valeur de l’appliquée BD. [Solution: ]…si l’on prend la difference du numérateur, & qu’on la divise par la difference du denominateur, apres avoir fait x = a = Ab ou AB, l’on aura la valeur cherchée de l’appliquée bd ou BD.” Translation : “Let there be a curve AMD (where AP = X, PM = y, AB = a) such that the value of the ordinate y is expressed by a fraction whose numerator and denominator each become zero when x = a ; that is, when the point P falls on the given point B. One asks what shall then be the value of the ordinate BD. [Solution: ]… if one takes the differential of the numerator and if one divides it by the differential of the denominator, after having set x = a = Ab or AB, one will have the value [that was] sought of the ordinate bd or BD.”
3. ^ Weisstein, Eric W., "L'Hospital's Rule" from MathWorld.
4. ^ Spivak, Michael (1994). Calculus. Houston, Texas: Publish or Perish. pp. 201–202, 210–211. ISBN 0-914098-89-6. 

External links

• l'Hôpital's rule at PlanetMath