L'Hôpital's rule

In mathematics, more specifically calculus, L'Hôpital's rule or L'Hospital's rule (French: [lopital]) provides a technique to evaluate limits of indeterminate forms. Application (or repeated application) of the rule often converts an indeterminate form to an expression that can be easily evaluated by substitution. The rule is named after the 17th-century French mathematician Guillaume de l'Hôpital. Although the rule is often attributed to L'Hôpital, the theorem was first introduced to him in 1694 by the Swiss mathematician Johann Bernoulli.

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 all of R by defining h(0) = f′(0)/g′(0) = −2.

L'Hôpital's rule states that for functions f and g which are differentiable on an open interval I except possibly at a point c contained in I, if ${\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty ,}$ ${\displaystyle g'(x)\neq 0}$ for all x in I with x ≠ c, and ${\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$ exists, then

${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}.}$

The differentiation of the numerator and denominator often simplifies the quotient or converts it to a limit that can be evaluated directly.

History

Guillaume de l'Hôpital (also written l'Hospital[lower-alpha 1]) published this rule in his 1696 book Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes (literal translation: Analysis of the Infinitely Small for the Understanding of Curved Lines), the first textbook on differential calculus.[1][lower-alpha 2] However, it is believed that the rule was discovered by the Swiss mathematician Johann Bernoulli.[3][4]

General form

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). Let I be an open interval containing c (for a two-sided limit) or an open interval with endpoint c (for a one-sided limit, or a limit at infinity if c is infinite). The real valued functions f and g are assumed to be differentiable on I except possibly at c, and additionally ${\displaystyle g'(x)\neq 0}$ on I except possibly at c. It is also assumed that ${\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}=L.}$ Thus the rule applies to situations in which the ratio of the derivatives has a finite or infinite limit, but not to situations in which that ratio fluctuates permanently as x gets closer and closer to c.

If either

${\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0}$

or

${\displaystyle \lim _{x\to c}|f(x)|=\lim _{x\to c}|g(x)|=\infty ,}$

then

${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=L.}$

Although we have written x → c throughout, the limits may also be one-sided limits (x → c⁺ or x → c⁻), when c is a finite endpoint of I.

In the second case, the hypothesis that f diverges 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 ${\displaystyle \lim _{x\to c}|g(x)|=\infty .}$

The hypothesis that ${\displaystyle g'(x)\neq 0}$ appears most commonly in the literature, but some authors sidestep this hypothesis by adding other hypotheses elsewhere. One method[5] is to define the limit of a function with the additional requirement that the limiting function is defined everywhere on the relevant interval I except possibly at c.[lower-alpha 3] Another method[6] is to require that both f and g be differentiable everywhere on an interval containing c.

Requirement that the limit exist

The requirement that the limit

${\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$

exists is essential. Without this condition, ${\displaystyle f'}$ or ${\displaystyle g'}$ may exhibit undamped oscillations as ${\displaystyle x}$ approaches ${\displaystyle c}$, in which case L'Hôpital's rule does not apply. For example, if ${\displaystyle f(x)=x+\sin(x)}$, ${\displaystyle g(x)=x}$ and ${\displaystyle c=\pm \infty }$, then

${\displaystyle {\frac {f'(x)}{g'(x)}}={\frac {1+\cos(x)}{1}};}$

this expression does not approach a limit as ${\displaystyle x}$ goes to ${\displaystyle c}$, since the cosine function oscillates between 1 and −1. But working with the original functions, ${\displaystyle \lim _{x\to \infty }{\frac {f(x)}{g(x)}}}$ can be shown to exist:

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

In a case such as this, all that can be concluded is that

${\displaystyle \liminf _{x\to c}{\frac {f'(x)}{g'(x)}}\leq \liminf _{x\to c}{\frac {f(x)}{g(x)}}\leq \limsup _{x\to c}{\frac {f(x)}{g(x)}}\leq \limsup _{x\to c}{\frac {f'(x)}{g'(x)}},}$

so that if the limit of f/g exists, then it must lie between the inferior and superior limits of f′/g′. (In the example above, this is true, since 1 indeed lies between 0 and 2.)

Examples

• Here is a basic example involving the exponential function, which involves the indeterminate form 0/0 at x = 0:

${\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {e^{x}-1}{x^{2}+x}}&=\lim _{x\to 0}{\frac {{\frac {d}{dx}}(e^{x}-1)}{{\frac {d}{dx}}(x^{2}+x)}}\\[4pt]&=\lim _{x\to 0}{\frac {e^{x}}{2x+1}}\\[4pt]&=1.\end{aligned}}}$

• 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:

${\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {2\sin(x)-\sin(2x)}{x-\sin(x)}}&=\lim _{x\to 0}{\frac {2\cos(x)-2\cos(2x)}{1-\cos(x)}}\\[4pt]&=\lim _{x\to 0}{\frac {-2\sin(x)+4\sin(2x)}{\sin(x)}}\\[4pt]&=\lim _{x\to 0}{\frac {-2\cos(x)+8\cos(2x)}{\cos(x)}}\\[4pt]&={\frac {-2+8}{1}}\\[4pt]&=6.\end{aligned}}}$

• Here is an example involving ∞/∞:

${\displaystyle \lim _{x\to \infty }x^{n}\cdot e^{-x}=\lim _{x\to \infty }{\frac {x^{n}}{e^{x}}}=\lim _{x\to \infty }{\frac {nx^{n-1}}{e^{x}}}=n\cdot \lim _{x\to \infty }{\frac {x^{n-1}}{e^{x}}}.}$

Repeatedly apply L'Hôpital's rule until the exponent is zero (if n is an integer) or negative (if n is fractional) to conclude that the limit is zero.

• Here is an example involving the indeterminate form 0 · ∞ (see below), which is rewritten as the form ∞/∞:

${\displaystyle \lim _{x\to 0^{+}}x\ln x=\lim _{x\to 0^{+}}{\frac {\ln x}{\frac {1}{x}}}=\lim _{x\to 0^{+}}{\frac {\frac {1}{x}}{-{\frac {1}{x^{2}}}}}=\lim _{x\to 0^{+}}-x=0.}$

• 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 ${\displaystyle {\frac {P}{n}}}$ (since only principal is being repaid); this is consistent with the formula for non-zero interest rates:

${\displaystyle {\begin{aligned}\lim _{r\to 0}{\frac {Pr(1+r)^{n}}{(1+r)^{n}-1}}&=P\lim _{r\to 0}{\frac {(1+r)^{n}+rn(1+r)^{n-1}}{n(1+r)^{n-1}}}\\[4pt]&={\frac {P}{n}}.\end{aligned}}}$

• One can also use L'Hôpital's rule to prove the following theorem. If f is twice-differentiable in a neighborhood of x, then

${\displaystyle {\begin{aligned}\lim _{h\to 0}{\frac {f(x+h)+f(x-h)-2f(x)}{h^{2}}}&=\lim _{h\to 0}{\frac {f'(x+h)-f'(x-h)}{2h}}\\[4pt]&=\lim _{h\to 0}{\frac {f''(x+h)+f''(x-h)}{2}}\\[4pt]&=f''(x).\end{aligned}}}$

• Sometimes L'Hôpital's rule is invoked in a tricky way: suppose f(x) + f′(x) converges as x → ∞ and that ${\displaystyle e^{x}\cdot f(x)}$ converges to positive or negative infinity. Then:

${\displaystyle \lim _{x\to \infty }f(x)=\lim _{x\to \infty }{\frac {e^{x}\cdot f(x)}{e^{x}}}=\lim _{x\to \infty }{\frac {e^{x}{\bigl (}f(x)+f'(x){\bigr )}}{e^{x}}}=\lim _{x\to \infty }{\bigl (}f(x)+f'(x){\bigr )}}$

and so,  ${\displaystyle \lim _{x\to \infty }f(x)}$  exists and  ${\displaystyle \lim _{x\to \infty }f'(x)=0.}$

The result remains true without the added hypothesis that ${\displaystyle e^{x}\cdot f(x)}$ converges to positive or negative infinity, but the justification is then incomplete.

Complications

Sometimes L'Hôpital's rule does not lead to an answer in a finite number of steps unless some additional steps are applied. Examples include the following:

• Two applications can lead to a return to the original expression that was to be evaluated:

${\displaystyle \lim _{x\to \infty }{\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}=\lim _{x\to \infty }{\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}=\lim _{x\to \infty }{\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}=\cdots .}$

This situation can be dealt with by substituting ${\displaystyle y=e^{x}}$ and noting that y goes to infinity as x goes to infinity; with this substitution, this problem can be solved with a single application of the rule:

${\displaystyle \lim _{x\to \infty }{\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}=\lim _{y\to \infty }{\frac {y+y^{-1}}{y-y^{-1}}}=\lim _{y\to \infty }{\frac {1-y^{-2}}{1+y^{-2}}}={\frac {1}{1}}=1.}$

Alternatively, the numerator and denominator can both be multiplied by ${\displaystyle e^{x},}$ at which point L'Hôpital's rule can immediately be applied successfully:[7]

${\displaystyle \lim _{x\to \infty }{\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}=\lim _{x\to \infty }{\frac {e^{2x}+1}{e^{2x}-1}}=\lim _{x\to \infty }{\frac {2e^{2x}}{2e^{2x}}}=1.}$

• An arbitrarily large number of applications may never lead to an answer even without repeating:

${\displaystyle \lim _{x\to \infty }{\frac {x^{\frac {1}{2}}+x^{-{\frac {1}{2}}}}{x^{\frac {1}{2}}-x^{-{\frac {1}{2}}}}}=\lim _{x\to \infty }{\frac {{\frac {1}{2}}x^{-{\frac {1}{2}}}-{\frac {1}{2}}x^{-{\frac {3}{2}}}}{{\frac {1}{2}}x^{-{\frac {1}{2}}}+{\frac {1}{2}}x^{-{\frac {3}{2}}}}}=\lim _{x\to \infty }{\frac {-{\frac {1}{4}}x^{-{\frac {3}{2}}}+{\frac {3}{4}}x^{-{\frac {5}{2}}}}{-{\frac {1}{4}}x^{-{\frac {3}{2}}}-{\frac {3}{4}}x^{-{\frac {5}{2}}}}}=\cdots .}$

This situation too can be dealt with by a transformation of variables, in this case ${\displaystyle y={\sqrt {x}}}$:

${\displaystyle \lim _{x\to \infty }{\frac {x^{\frac {1}{2}}+x^{-{\frac {1}{2}}}}{x^{\frac {1}{2}}-x^{-{\frac {1}{2}}}}}=\lim _{y\to \infty }{\frac {y+y^{-1}}{y-y^{-1}}}=\lim _{y\to \infty }{\frac {1-y^{-2}}{1+y^{-2}}}={\frac {1}{1}}=1.}$

Again, an alternative approach is to multiply numerator and denominator by ${\displaystyle x^{1/2}}$ before applying L'Hôpital's rule:

${\displaystyle \lim _{x\to \infty }{\frac {x^{\frac {1}{2}}+x^{-{\frac {1}{2}}}}{x^{\frac {1}{2}}-x^{-{\frac {1}{2}}}}}=\lim _{x\to \infty }{\frac {x+1}{x-1}}=\lim _{x\to \infty }{\frac {1}{1}}=1.}$

A common pitfall is using L'Hôpital's rule with some circular reasoning to compute a derivative via a difference quotient. For example, consider the task of proving the derivative formula for powers of x:

${\displaystyle \lim _{h\to 0}{\frac {(x+h)^{n}-x^{n}}{h}}=nx^{n-1}.}$

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.

Counterexamples when the derivative of the denominator is zero

The necessity of the condition that ${\displaystyle g'(x)\neq 0}$ near ${\displaystyle c}$ can be seen by the following counterexample due to Otto Stolz.[8] Let ${\displaystyle f(x)=x+\sin x\cos x}$ and ${\displaystyle g(x)=f(x)e^{\sin x}.}$ Then there is no limit for ${\displaystyle f(x)/g(x)}$ as ${\displaystyle x\to \infty .}$ However,

${\displaystyle {\begin{aligned}{\frac {f'(x)}{g'(x)}}&={\frac {2\cos ^{2}x}{(2\cos ^{2}x)e^{\sin x}+(x+\sin x\cos x)e^{\sin x}\cos x}}\\[4pt]&={\frac {2\cos x}{2\cos x+x+\sin x\cos x}}e^{-\sin x},\end{aligned}}}$

which tends to 0 as ${\displaystyle x\to \infty }$. Further examples of this type were found by Ralph P. Boas Jr.[9]

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:

${\displaystyle {\begin{aligned}\lim _{x\to 1}\left({\frac {x}{x-1}}-{\frac {1}{\ln x}}\right)&=\lim _{x\to 1}{\frac {x\cdot \ln x-x+1}{(x-1)\cdot \ln x}}&\quad (1)\\[6pt]&=\lim _{x\to 1}{\frac {\ln x}{{\frac {x-1}{x}}+\ln x}}&\quad (2)\\[6pt]&=\lim _{x\to 1}{\frac {x\cdot \ln x}{x-1+x\cdot \ln x}}&\quad (3)\\[6pt]&=\lim _{x\to 1}{\frac {1+\ln x}{1+1+\ln x}}&\quad (4)\\[6pt]&=\lim _{x\to 1}{\frac {1+\ln x}{2+\ln x}}\\[6pt]&={\frac {1}{2}},\end{aligned}}}$

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:

${\displaystyle \lim _{x\to 0^{+}}x^{x}=\lim _{x\to 0^{+}}e^{\ln(x^{x})}=\lim _{x\to 0^{+}}e^{x\cdot \ln x}=e^{\lim \limits _{x\to 0^{+}}(x\cdot \ln x)}.}$

It is valid to move the limit inside the exponential function because the exponential function is continuous. Now the exponent ${\displaystyle x}$ has been "moved down". The limit ${\displaystyle \lim _{x\to 0^{+}}x\cdot \ln x}$ is of the indeterminate form 0 · ∞, but as shown in an example above, l'Hôpital's rule may be used to determine that

${\displaystyle \lim _{x\to 0^{+}}x\cdot \ln x=0.}$

Thus

${\displaystyle \lim _{x\to 0^{+}}x^{x}=e^{0}=1.}$

Stolz–Cesàro theorem

Main article: Stolz–Cesàro theorem

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

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), 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 curve 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.

Proof of L'Hôpital's rule

Special case

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 ${\displaystyle f(c)=g(c)=0}$, and that ${\displaystyle g'(c)\neq 0}$. Then

${\displaystyle {\begin{aligned}&\lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f(x)-0}{g(x)-0}}=\lim _{x\to c}{\frac {f(x)-f(c)}{g(x)-g(c)}}\\[6pt]={}&\lim _{x\to c}{\frac {\left({\frac {f(x)-f(c)}{x-c}}\right)}{\left({\frac {g(x)-g(c)}{x-c}}\right)}}={\frac {\lim \limits _{x\to c}\left({\frac {f(x)-f(c)}{x-c}}\right)}{\lim \limits _{x\to c}\left({\frac {g(x)-g(c)}{x-c}}\right)}}={\frac {f'(c)}{g'(c)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}.\end{aligned}}}$

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 ${\displaystyle g'(c)\neq 0}$.

The proof of a more general version of L'Hôpital's rule is given below.

General proof

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 ${\displaystyle {\mathcal {I}}}$ be the open interval in the hypothesis with endpoint c. Considering that ${\displaystyle g'(x)\neq 0}$ on this interval and g is continuous, ${\displaystyle {\mathcal {I}}}$ can be chosen smaller so that g is nonzero on ${\displaystyle {\mathcal {I}}}$.[lower-alpha 4]

For each x in the interval, define ${\displaystyle m(x)=\inf {\frac {f'(\xi )}{g'(\xi )}}}$ and ${\displaystyle M(x)=\sup {\frac {f'(\xi )}{g'(\xi )}}}$ as ${\displaystyle \xi }$ 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 ${\displaystyle {\mathcal {I}}}$, Cauchy's mean value theorem ensures that for any two distinct points x and y in ${\displaystyle {\mathcal {I}}}$ there exists a ${\displaystyle \xi }$ between x and y such that ${\displaystyle {\frac {f(x)-f(y)}{g(x)-g(y)}}={\frac {f'(\xi )}{g'(\xi )}}}$. Consequently, ${\displaystyle m(x)\leq {\frac {f(x)-f(y)}{g(x)-g(y)}}\leq M(x)}$ 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.

Case 1: ${\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0}$

For any x in the interval ${\displaystyle {\mathcal {I}}}$, and point y between x and c,

${\displaystyle m(x)\leq {\frac {f(x)-f(y)}{g(x)-g(y)}}={\frac {{\frac {f(x)}{g(x)}}-{\frac {f(y)}{g(x)}}}{1-{\frac {g(y)}{g(x)}}}}\leq M(x)}$

and therefore as y approaches c, ${\displaystyle {\frac {f(y)}{g(x)}}}$ and ${\displaystyle {\frac {g(y)}{g(x)}}}$ become zero, and so

${\displaystyle m(x)\leq {\frac {f(x)}{g(x)}}\leq M(x).}$

Case 2: ${\displaystyle \lim _{x\to c}|g(x)|=\infty }$

For every x in the interval ${\displaystyle {\mathcal {I}}}$, define ${\displaystyle S_{x}=\{y\mid y{\text{ is between }}x{\text{ and }}c\}}$. For every point y between x and c,

${\displaystyle m(x)\leq {\frac {f(y)-f(x)}{g(y)-g(x)}}={\frac {{\frac {f(y)}{g(y)}}-{\frac {f(x)}{g(y)}}}{1-{\frac {g(x)}{g(y)}}}}\leq M(x).}$

As y approaches c, both ${\displaystyle {\frac {f(x)}{g(y)}}}$ and ${\displaystyle {\frac {g(x)}{g(y)}}}$ become zero, and therefore

${\displaystyle m(x)\leq \liminf _{y\in S_{x}}{\frac {f(y)}{g(y)}}\leq \limsup _{y\in S_{x}}{\frac {f(y)}{g(y)}}\leq M(x).}$

The limit superior and limit inferior are necessary since the existence of the limit of f/g has not yet been established.

It is also the case that

${\displaystyle \lim _{x\to c}m(x)=\lim _{x\to c}M(x)=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}=L.}$

[lower-alpha 5] and

${\displaystyle \lim _{x\to c}\left(\liminf _{y\in S_{x}}{\frac {f(y)}{g(y)}}\right)=\liminf _{x\to c}{\frac {f(x)}{g(x)}}}$ and ${\displaystyle \lim _{x\to c}\left(\limsup _{y\in S_{x}}{\frac {f(y)}{g(y)}}\right)=\limsup _{x\to c}{\frac {f(x)}{g(x)}}.}$

In case 1, the squeeze theorem establishes that ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}}$ exists and is equal to L. In the case 2, and the squeeze theorem again asserts that ${\displaystyle \liminf _{x\to c}{\frac {f(x)}{g(x)}}=\limsup _{x\to c}{\frac {f(x)}{g(x)}}=L}$, and so the limit ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}}$ exists and is equal to L. This is the result that was to be proven.

In case 2 the assumption that f(x) diverges to infinity was not used 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.[10]

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.

Corollary

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 ${\displaystyle f'(x)}$ exists for all x in some open interval containing a, except perhaps for ${\displaystyle x=a}$. Suppose, moreover, that ${\displaystyle \lim _{x\to a}f'(x)}$ exists. Then ${\displaystyle f'(a)}$ also exists and

${\displaystyle f'(a)=\lim _{x\to a}f'(x).}$

In particular, f' is also continuous at a.

Proof

Consider the functions ${\displaystyle h(x)=f(x)-f(a)}$ and ${\displaystyle g(x)=x-a}$. The continuity of f at a tells us that ${\displaystyle \lim _{x\to a}h(x)=0}$. Moreover, ${\displaystyle \lim _{x\to a}g(x)=0}$ since a polynomial function is always continuous everywhere. Applying L'Hopital's rule shows that ${\displaystyle f'(a):=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}=\lim _{x\to a}{\frac {h(x)}{g(x)}}=\lim _{x\to a}f'(x)}$.

See also

• L'Hôpital controversy

Notes

1. 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.
2. "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."[2]
3. The functional analysis definition of the limit of a function does not require the existence of such an interval.
4. 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.
5. The limits ${\displaystyle \lim _{x\to c}m(x)}$ and ${\displaystyle \lim _{x\to c}M(x)}$ both exist as they feature nondecreasing and nonincreasing functions of x, respectively. Consider a sequence ${\displaystyle x_{i}\to c}$. Then ${\displaystyle \lim _{i}m(x_{i})\leq \lim _{i}{\frac {f'(x_{i})}{g'(x_{i})}}\leq \lim _{i}M(x_{i})}$, as the inequality holds for each i; this yields the inequalities ${\displaystyle \lim _{x\to c}m(x)\leq \lim _{x\to c}{\frac {f'(x)}{g'(x)}}\leq \lim _{x\to c}M(x)}$ The next step is to show ${\displaystyle \lim _{x\to c}M(x)\leq \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$. Indeed, fix a sequence of numbers ${\displaystyle \varepsilon _{i}>0}$ such that ${\displaystyle \lim _{i}\varepsilon _{i}=0}$, and a sequence ${\displaystyle x_{i}\to c}$. For each i, choose ${\displaystyle x_{i}<y_{i}<c}$ such that ${\displaystyle {\frac {f'(y_{i})}{g'(y_{i})}}+\varepsilon _{i}\geq \sup _{x_{i}<\xi <c}{\frac {f'(\xi )}{g'(\xi )}}}$, by the definition of ${\displaystyle \sup }$. Thus ${\displaystyle \lim _{i}M(x_{i})\leq \lim _{i}{\frac {f'(y_{i})}{g'(y_{i})}}+\varepsilon _{i}=\lim _{i}{\frac {f'(y_{i})}{g'(y_{i})}}+\lim _{i}\varepsilon _{i}=\lim _{i}{\frac {f'(y_{i})}{g'(y_{i})}}}$ as desired. The argument that ${\displaystyle \lim _{x\to c}m(x)\geq \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$ is similar.

References

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. Retrieved 21 December 2008.
2. L’Hospital. "Analyse des infiniment petits": 145–146. Cite journal requires |journal= (help)
3. Boyer, Carl B.; Merzbach, Uta C. (2011). A History of Mathematics (3rd illustrated ed.). John Wiley & Sons. p. 321. ISBN 978-0-470-63056-3. Extract of page 321
4. Weisstein, Eric W. "L'Hospital's Rule". MathWorld.
5. (Chatterjee 2005, p. 291)
6. (Krantz 2004, p.79)
7. Multiplying by ${\displaystyle e^{-}x}$ instead yields a solution to the limit without need for l'Hôpital's rule.
8. Stolz, Otto (1879). "Ueber die Grenzwerthe der Quotienten" [About the limits of quotients]. Mathematische Annalen (in German). 15: 556–559. doi:10.1007/bf02086277.
9. Boas Jr., Ralph P. (1986). "Counterexamples to L'Hopital's Rule". American Mathematical Monthly. 93 (8): 644–645. doi:10.1080/00029890.1986.11971912. JSTOR 2322330.
10. "L'Hopital's Theorem". IMOmath. International Mathematical Olympiad.

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