Example application of l'Hôpital's rule to f(x)=sin(x) and g(x)=-0.5x: the function h(x) = f(x)/g(x) is undefined at x=0, but can be completed to a continuous function on whole ℝ by defining h(0)=f’(0)/g’(0)=-2.
The general form of L'Hôpital's rule covers many cases. Let c and L be extended real numbers (i.e., real numbers, positive infinity, or negative infinity). The real valued functions f and g are assumed to be differentiable on an open interval with endpoint c, and additionally on the interval. It is also assumed that Thus the rule applies to situations in which the ratio of the derivatives has a finite or infinite limit, and not to situations in which that ratio fluctuates permanently as x gets closer and closer to c.
The limits may also be one-sided limits. In the second case, the hypothesis that fdiverges to infinity is not used in the proof (see note at the end of the proof section); thus, while the conditions of the rule are normally stated as above, the second sufficient condition for the rule's procedure to be valid can be more briefly stated as
The hypothesis "" appears most commonly in the literature. Some authors sidestep this hypothesis by adding other hypotheses elsewhere. One method is to define the limit of a function with the additional requirement that the limiting function is defined everywhere on a connected interval with endpoint c. Another method is to require that both f and g be differentiable everywhere on an interval containing c.
Here is an example involving the sinc function, , which handles the indeterminate form 0/0 at x = 0:
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:
This example involves 0/0. Suppose that b > 0. Then
Here is another example involving 0/0:
This example involves ∞/∞. Assume n is a positive integer. Then
Repeatedly apply L'Hôpital's rule until the exponent is zero to conclude that the limit is zero.
Here is an example involving the Mortgage repayment formula and 0/0. Let P be the principal (loan amount), r the interest rate per period and n the number of periods. When r is zero, the repayment amount per period is (since only principal is being repaid); this is consistent with the formula for non-zero interest rates:
One can also use L'Hôpital's rule to prove the following theorem. If f′′ is continuous at x, then
Sometimes L'Hôpital's rule is invoked in a tricky way: suppose f(x) + f′(x) converges as x → ∞ and that converges to positive or negative infinity. Then:
and so, exists and
The result remains true without the added hypothesis that converges to positive or negative infinity, but the justification is then incomplete.
with , defined over , is said to have constant relative risk aversion equal to . But unit relative risk aversion cannot be expressed directly with this expression, since as approaches 1 the numerator and denominator both approach zero. However, a single application of L'Hôpital's rule allows this case to be expressed as
Applying L'Hôpital's rule and finding the derivatives with respect to h of the numerator and the denominator yields n xn−1 as expected. However, differentiating the numerator required the use of the very fact that is being proven. This is an example of begging the question, since one may not assume the fact to be proven during the course of the proof.
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:
where L'Hôpital's rule is applied when going from (1) to (2) and again when 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:
It is valid to move the limit inside the exponential function because the exponential function is continuous. Now the exponent has been "moved down". The limit is of the indeterminate form 0 × (−∞), but as shown in an example above, l'Hôpital's rule may be used to determine that
Consider the curve in the plane whose x-coordinate is given by g(t) and whose y-coordinate is given by f(t), with both functions continuous, i.e., the locus of points of the form [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 the tangent to the curve at the point [g(c), f(c)] = [0,0]. The tangent to the curve at the point [g(t), f(t)] is given by [g′(t), f′(t)]. L'Hôpital's rule then states that the slope of the tangent when t = c is the limit of the slope of the tangent to the curve as the curve approaches the origin, provided that this is defined.
The proof of L'Hôpital's rule is simple in the case where f and g are continuously differentiable at the point c and where a finite limit is found after the first round of differentiation. It is not a proof of the general L'Hôpital's rule because it is stricter in its definition, requiring both differentiability and that c be a real number. Since many common functions have continuous derivatives (e.g. polynomials, sine and cosine, exponential functions), it is a special case worthy of attention.
Suppose that f and g are continuously differentiable at a real number c, that , and that . Then
This follows from the difference-quotient definition of the derivative. The last equality follows from the continuity of the derivatives at c. The limit in the conclusion is not indeterminate because .
The proof of a more general version of L'Hôpital's rule is given below.
The following proof is due to (Taylor 1952), where a unified proof for the 0/0 and ±∞/±∞ indeterminate forms is given. Taylor notes that different proofs may be found in (Lettenmeyer 1936) and (Wazewski 1949).
Let f and g be functions satisfying the hypotheses in the General form section. Let be the open interval in the hypothesis with endpoint c. Considering that on this interval and g is continuous, can be chosen smaller so that g is nonzero on .
For each x in the interval, define and as ranges over all values between x and c. (The symbols inf and sup denote the infimum and supremum.)
From the differentiability of f and g on , Cauchy's mean value theorem ensures that for any two distinct points x and y in there exists a between x and y such that . Consequently for all choices of distinct x and y in the interval. The value g(x)-g(y) is always nonzero for distinct x and y in the interval, for if it was not, the mean value theorem would imply the existence of a p between x and y such that g' (p)=0.
The definition of m(x) and M(x) will result in an extended real number, and so it is possible for them to take on the values ±∞. In the following two cases, m(x) and M(x) will establish bounds on the ratio f/g.
For any x in the interval , and point y between x and c,
and therefore as y approaches c, and become zero, and so
For every x in the interval , define . For every point y between x and c, we have
As y approaches c, both and become zero, and therefore
In case 1, the squeeze theorem, establishes that exists and is equal to L. In the case 2, and the squeeze theorem again asserts that , and so the limit exists and is equal to L. This is the result that was to be proven.
Note: In case 2 we did not use the assumption that f(x) diverges to infinity within the proof. This means that if |g(x)| diverges to infinity as x approaches c and both f and g satisfy the hypotheses of L'Hôpital's rule, then no additional assumption is needed about the limit of f(x): It could even be the case that the limit of f(x) does not exist. In this case, L'Hopital's theorem is actually a consequence of Cesàro–Stolz (see proof at http://www.imomath.com/index.php?options=686).
In the case when |g(x)| diverges to infinity as x approaches c and f(x) converges to a finite limit at c, then L'Hôpital's rule would be applicable, but not absolutely necessary, since basic limit calculus will show that the limit of f(x)/g(x) as x approaches c must be zero.
A simple but very useful consequence of L'Hopital's rule is a well-known criterion for differentiability. It states the following: suppose that f is continuous at a, and that exists for all x in some interval containing a, except perhaps for . Suppose, moreover, that exists. Then also exists and
^In the 17th and 18th centuries, the name was commonly spelled "l'Hospital", and he himself spelled his name that way. However, French spellings have been altered: the silent 's' has been removed and replaced with the circumflex over the preceding vowel. The former spelling is still used in English where there is no circumflex.
^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. Retrieved 21 December 2008.
^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."
^Since g' is nonzero and g is continuous on the interval, it is impossible for g to be zero more than once on the interval. If it had two zeros, the mean value theorem would assert the existence of a point p in the interval between the zeros such that g' (p) = 0. So either g is already nonzero on the interval, or else the interval can be reduced in size so as not to contain the single zero of g.
^Note and exist as they are nondecreasing and nonincreasing functions of x, respectively. Consider a sequence , we easily have , as the inequality holds for each i; this yields the inequalities We show . Indeed, fix a sequence of numbers such that , and a sequence . For each i, we may choose such that , by the definition of . Thus we have as desired. The argument that is similar.