L'Hôpital's rule

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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 (/ˌloʊpiːˈtɑːl/, loh-pee-TAHL), also known as Bernoulli's rule, is a mathematical theorem that allows evaluating limits of indeterminate forms using derivatives. 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.

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 $\underset{x\to c}{lim}f(x)=\underset{x\to c}{lim}g(x)=0\text{ or }\pm \mathrm{\infty },$ and $g^{\prime }(x)\ne 0$ for all x in I with x ≠ c, and $\underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ exists, then

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

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) 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. However, it is believed that the rule was discovered by the Swiss mathematician Johann Bernoulli.

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 ${\displaystyle g^{\prime }(x)\ne 0}$ on I except possibly at c. It is also assumed that $\underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=L.$ 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

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

or

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

then

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

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 $\underset{x\to c}{lim}|g(x)|=\mathrm{\infty }.$

The hypothesis that ${\displaystyle g^{\prime }(x)\ne 0}$ appears most commonly in the literature, but 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 the relevant interval I except possibly at c. Another method is to require that both f and g be differentiable everywhere on an interval containing c.

Cases where theorem cannot be applied (Necessity of conditions)

All four conditions for L'Hôpital's rule are necessary:

1. Indeterminacy of form: ${\displaystyle \underset{x\to c}{lim}f(x)=\underset{x\to c}{lim}g(x)=0}$ or ${\displaystyle \pm \mathrm{\infty }}$ ; and
2. Differentiability of functions: $f(x)$ and $g(x)$ are differentiable on an open interval $\mathcal{I}$ except possibly at a point $c$ contained in $\mathcal{I}$ (the same point from the limit) ; and
3. Non-zero derivative of denominator: ${\displaystyle g^{\prime }(x)\ne 0}$ for all $x$ in $\mathcal{I}$ with ${\displaystyle x\ne c}$ ; and
4. Existence of limit of the quotient of the derivatives: ${\displaystyle \underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}}$ exists.

Where one of the above conditions is not satisfied, L'Hôpital's rule is not valid in general, and so it cannot always be applied.

Form is not indeterminate

The necessity of the first condition can be seen by considering the counterexample where the functions are ${\displaystyle f(x)=x+1}$ and ${\displaystyle g(x)=2x+1}$ and the limit is ${\displaystyle x\to 1}$.

The first condition is not satisfied for this counterexample because ${\displaystyle \underset{x\to 1}{lim}f(x)=\underset{x\to 1}{lim}(x+1)=(1)+1=2\ne 0}$ and ${\displaystyle \underset{x\to 1}{lim}g(x)=\underset{x\to 1}{lim}(2x+1)=2(1)+1=3\ne 0}$. This means that the form is not indeterminate.

The second and third conditions are satisfied by $f(x)$ and $g(x)$. The fourth condition is also satisfied with ${\displaystyle \underset{x\to 1}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=\underset{x\to 1}{lim}\frac{(x+1{)}^{\prime }}{(2x+1{)}^{\prime }}=\underset{x\to 1}{lim}\frac{1}{2}=\frac{1}{2}}$.

But, L'Hôpital's rule fails in this counterexample, since ${\displaystyle \underset{x\to 1}{lim}\frac{f(x)}{g(x)}=\underset{x\to 1}{lim}\frac{x+1}{2x+1}=\frac{\underset{x\to 1}{lim}(x+1)}{\underset{x\to 1}{lim}(2x+1)}=\frac{2}{3}\ne \frac{1}{2}=\underset{x\to 1}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}}$.

Differentiability of functions

Differentiability of functions is a requirement because if a function is not differentiable, then the derivative of the functions is not guaranteed to exist at each point in $\mathcal{I}$. The fact that $\mathcal{I}$ is an open interval is grandfathered in from the hypothesis of the Cauchy Mean Value Theorem. The notable exception of the possibility of the functions being not differentiable at $c$ exists because L'Hôpital's rule only requires the derivative to exist as the function approaches $c$; the derivative does not need to be taken at $c$.

For example, let ${\displaystyle f(x)=\{\begin{array}{ll}\sin x,& x\ne 0\\ 1,& x=0\end{array}}$ , ${\displaystyle g(x)=x}$, and ${\displaystyle c=0}$. In this case, $f(x)$ is not differentiable at $c$. However, since $f(x)$ is differentiable everywhere except $c$, then ${\displaystyle \underset{x\to c}{lim}{f}^{\prime }(x)}$ still exists. Thus, since

${\displaystyle \underset{x\to c}{lim}\frac{f(x)}{g(x)}=\frac{0}{0}}$ and ${\displaystyle \underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}}$ exists, L'Hôpital's rule still holds.

Derivative of denominator is zero

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

${\displaystyle \begin{array}{rl}\frac{f^{\prime }(x)}{g^{\prime }(x)}& =\frac{2{\cos }^{2}x}{(2{\cos }^{2}x)e^{\sin x}+(x+\sin x\cos x)e^{\sin x}\cos x}\\ & =\frac{2\cos x}{2\cos x+x+\sin x\cos x}{e}^{-\sin x},\end{array}}$

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

Limit of derivatives does not exist

The requirement that the limit

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

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

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

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

${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{f(x)}{g(x)}=\underset{x\to \mathrm{\infty }}{lim}\left(\frac{x+\sin (x)}{x}\right)=\underset{x\to \mathrm{\infty }}{lim}\left(1+\frac{\sin (x)}{x}\right)=1+\underset{x\to \mathrm{\infty }}{lim}\left(\frac{\sin (x)}{x}\right)=1+0=1.}$

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

${\displaystyle \underset{x\to c}{lim inf}\frac{f^{\prime }(x)}{g^{\prime }(x)}\le \underset{x\to c}{lim inf}\frac{f(x)}{g(x)}\le \underset{x\to c}{lim sup}\frac{f(x)}{g(x)}\le \underset{x\to c}{lim sup}\frac{f^{\prime }(x)}{g^{\prime }(x)},}$

so that if the limit of $\frac{f}{g}$ f/g exists, then it must lie between the inferior and superior limits of $\frac{f^{\prime }}{g^{\prime }}$ . (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:
  $${\displaystyle \begin{array}{rl}\underset{x\to 0}{lim}\frac{e^x-1}{x^2+x}& =\underset{x\to 0}{lim}\frac{\frac{d}{dx}(e^x-1)}{\frac{d}{dx}(x^2+x)}\\ & =\underset{x\to 0}{lim}\frac{e^x}{2x+1}\\ & =1.\end{array}}$$

  
• 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:
  $${\displaystyle \begin{array}{rl}\underset{x\to 0}{lim}\frac{2\sin (x)-\sin (2x)}{x-\sin (x)}& =\underset{x\to 0}{lim}\frac{2\cos (x)-2\cos (2x)}{1-\cos (x)}\\ & =\underset{x\to 0}{lim}\frac{-2\sin (x)+4\sin (2x)}{\sin (x)}\\ & =\underset{x\to 0}{lim}\frac{-2\cos (x)+8\cos (2x)}{\cos (x)}\\ & =\frac{-2+8}{1}\\ & =6.\end{array}}$$

  
• Here is an example involving ∞/∞:
  $${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{x}^n\cdot e^{-x}=\underset{x\to \mathrm{\infty }}{lim}\frac{x^n}{e^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{n{x}^{n-1}}{e^x}=n\cdot \underset{x\to \mathrm{\infty }}{lim}\frac{x^{n-1}}{e^x}.}$$

  
  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 ∞/∞:
  $${\displaystyle \underset{x\to 0^{+}}{lim}x\ln x=\underset{x\to 0^{+}}{lim}\frac{\ln x}{\frac{1}{x}}=\underset{x\to 0^{+}}{lim}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\underset{x\to 0^{+}}{lim}-x=0.}$$

  
• 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:
  $${\displaystyle \begin{array}{rl}\underset{r\to 0}{lim}\frac{Pr(1+r{)}^n}{(1+r{)}^n-1}& =P\underset{r\to 0}{lim}\frac{(1+r{)}^n+rn(1+r{)}^{n-1}}{n(1+r{)}^{n-1}}\\ & =\frac{P}{n}.\end{array}}$$

  
• One can also use L'Hôpital's rule to prove the following theorem. If f is twice-differentiable in a neighborhood of x and that its second derivative is continuous on this neighbourhood, then
  $${\displaystyle \begin{array}{rl}\underset{h\to 0}{lim}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}& =\underset{h\to 0}{lim}\frac{f^{\prime }(x+h)-f^{\prime }(x-h)}{2h}\\ & =\underset{h\to 0}{lim}\frac{f^{″}(x+h)+f^{″}(x-h)}{2}\\ & =f^{″}(x).\end{array}}$$

  

• 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:

  $${\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)=\underset{x\to \mathrm{\infty }}{lim}\frac{e^x\cdot f(x)}{e^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{e^x(f(x)+f^{\prime }(x))}{e^x}=\underset{x\to \mathrm{\infty }}{lim}(f(x)+f^{\prime }(x))}$$

  
  and so, $\underset{x\to \mathrm{\infty }}{lim}f(x)$ exists and $\underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(x)=0.$

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

  
  This situation can be dealt with by substituting $y=e^x$ 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:
  $${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{e^x+e^{-x}}{e^x-e^{-x}}=\underset{y\to \mathrm{\infty }}{lim}\frac{y+y^{-1}}{y-y^{-1}}=\underset{y\to \mathrm{\infty }}{lim}\frac{1-y^{-2}}{1+y^{-2}}=\frac{1}{1}=1.}$$

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

  
• An arbitrarily large number of applications may never lead to an answer even without repeating:
  $${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{x^{\frac{1}{2}}+x^{-\frac{1}{2}}}{x^{\frac{1}{2}}-x^{-\frac{1}{2}}}=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{2}{x}^{-\frac{1}{2}}-\frac{1}{2}{x}^{-\frac{3}{2}}}{\frac{1}{2}{x}^{-\frac{1}{2}}+\frac{1}{2}{x}^{-\frac{3}{2}}}=\underset{x\to \mathrm{\infty }}{lim}\frac{-\frac{1}{4}{x}^{-\frac{3}{2}}+\frac{3}{4}{x}^{-\frac{5}{2}}}{-\frac{1}{4}{x}^{-\frac{3}{2}}-\frac{3}{4}{x}^{-\frac{5}{2}}}=\cdots .}$$

  
  This situation too can be dealt with by a transformation of variables, in this case $y=\sqrt{x}$:
  $${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{x^{\frac{1}{2}}+x^{-\frac{1}{2}}}{x^{\frac{1}{2}}-x^{-\frac{1}{2}}}=\underset{y\to \mathrm{\infty }}{lim}\frac{y+y^{-1}}{y-y^{-1}}=\underset{y\to \mathrm{\infty }}{lim}\frac{1-y^{-2}}{1+y^{-2}}=\frac{1}{1}=1.}$$

  
  Again, an alternative approach is to multiply numerator and denominator by $x^{1/2}$ before applying L'Hôpital's rule:
  $${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{x^{\frac{1}{2}}+x^{-\frac{1}{2}}}{x^{\frac{1}{2}}-x^{-\frac{1}{2}}}=\underset{x\to \mathrm{\infty }}{lim}\frac{x+1}{x-1}=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{1}=1.}$$

  

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:

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

Applying L'Hôpital's rule and finding the derivatives with respect to h of the numerator and the denominator yields nxⁿ⁻¹ as expected. However, differentiating the numerator requires 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.

A similar pitfall occurs in the calculation of ${\displaystyle \underset{x\to 0}{lim}\frac{\sin (x)}{x}=1.}$ Proving that differentiating $\sin (x)$ gives $\cos (x)$ involves calculating the difference quotient ${\displaystyle \underset{h\to 0}{lim}\frac{\sin (h)}{h}}$ in the first place, so a different method such as squeeze theorem must be used instead.

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:

${\displaystyle \begin{array}{rlr}\underset{x\to 1}{lim}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)& =\underset{x\to 1}{lim}\frac{x\cdot \ln x-x+1}{(x-1)\cdot \ln x}& (1)\\ & =\underset{x\to 1}{lim}\frac{\ln x}{\frac{x-1}{x}+\ln x}& (2)\\ & =\underset{x\to 1}{lim}\frac{x\cdot \ln x}{x-1+x\cdot \ln x}& (3)\\ & =\underset{x\to 1}{lim}\frac{1+\ln x}{1+1+\ln x}& (4)\\ & =\underset{x\to 1}{lim}\frac{1+\ln x}{2+\ln x}\\ & =\frac{1}{2},\end{array}}$

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:

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

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 ${\displaystyle \underset{x\to 0^{+}}{lim}x\cdot \ln x}$ is of the indeterminate form 0 · ∞, but as shown in an example above, l'Hôpital's rule may be used to determine that

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

Thus

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

The following table lists the most common indeterminate forms, and the transformations for applying l'Hôpital's rule:

Indeterminate form	Conditions	Transformation to $0/0$
0/0	$\underset{x\to c}{lim}f(x)=0,\underset{x\to c}{lim}g(x)=0$	—
$\mathrm{\infty }$/$\mathrm{\infty }$	$\underset{x\to c}{lim}f(x)=\mathrm{\infty },\underset{x\to c}{lim}g(x)=\mathrm{\infty }$	$\underset{x\to c}{lim}\frac{f(x)}{g(x)}=\underset{x\to c}{lim}\frac{1/g(x)}{1/f(x)}$
$0\cdot \mathrm{\infty }$	$\underset{x\to c}{lim}f(x)=0,\underset{x\to c}{lim}g(x)=\mathrm{\infty }$	$\underset{x\to c}{lim}f(x)g(x)=\underset{x\to c}{lim}\frac{f(x)}{1/g(x)}$
$\mathrm{\infty }-\mathrm{\infty }$	$\underset{x\to c}{lim}f(x)=\mathrm{\infty },\underset{x\to c}{lim}g(x)=\mathrm{\infty }$	$\underset{x\to c}{lim}(f(x)-g(x))=\underset{x\to c}{lim}\frac{1/g(x)-1/f(x)}{1/(f(x)g(x))}$
$0^0$	$\underset{x\to c}{lim}f(x)=0^{+},\underset{x\to c}{lim}g(x)=0$	$\underset{x\to c}{lim}f(x{)}^{g(x)}=\exp \underset{x\to c}{lim}\frac{g(x)}{1/\ln f(x)}$
$1^{\mathrm{\infty }}$	$\underset{x\to c}{lim}f(x)=1,\underset{x\to c}{lim}g(x)=\mathrm{\infty }$	$\underset{x\to c}{lim}f(x{)}^{g(x)}=\exp \underset{x\to c}{lim}\frac{\ln f(x)}{1/g(x)}$
${\displaystyle {\mathrm{\infty }}^0}$	$\underset{x\to c}{lim}f(x)=\mathrm{\infty },\underset{x\to c}{lim}g(x)=0$	$\underset{x\to c}{lim}f(x{)}^{g(x)}=\exp \underset{x\to c}{lim}\frac{g(x)}{1/\ln f(x)}$

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 $g^{\prime }(c)\ne 0$. Then

${\displaystyle \begin{array}{rl}& \underset{x\to c}{lim}\frac{f(x)}{g(x)}=\underset{x\to c}{lim}\frac{f(x)-0}{g(x)-0}=\underset{x\to c}{lim}\frac{f(x)-f(c)}{g(x)-g(c)}\\ =& \underset{x\to c}{lim}\frac{\left(\frac{f(x)-f(c)}{x-c}\right)}{\left(\frac{g(x)-g(c)}{x-c}\right)}=\frac{\underset{x\to c}{lim}\left(\frac{f(x)-f(c)}{x-c}\right)}{\underset{x\to c}{lim}\left(\frac{g(x)-g(c)}{x-c}\right)}=\frac{f^{\prime }(c)}{g^{\prime }(c)}=\underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}.\end{array}}$

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 ${\displaystyle g^{\prime }(c)\ne 0}$.

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 $\frac{0}{0}$ and $\frac{\pm \mathrm{\infty }}{\pm \mathrm{\infty }}$ 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 $\mathcal{I}$ be the open interval in the hypothesis with endpoint c. Considering that ${\displaystyle g^{\prime }(x)\ne 0}$ on this interval and g is continuous, $\mathcal{I}$ can be chosen smaller so that g is nonzero on $\mathcal{I}$.^[d]

For each x in the interval, define $m(x)=inf\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}$ and $M(x)=sup\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}$ as $\xi$ 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 $\mathcal{I}$, Cauchy's mean value theorem ensures that for any two distinct points x and y in $\mathcal{I}$ there exists a $\xi$ between x and y such that $\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}$. Consequently, $m(x)\le \frac{f(x)-f(y)}{g(x)-g(y)}\le M(x)$ 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: ${\displaystyle \underset{x\to c}{lim}f(x)=\underset{x\to c}{lim}g(x)=0}$

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

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

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

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

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

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

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

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

${\displaystyle m(x)\le \underset{y\in S_x}{lim inf}\frac{f(y)}{g(y)}\le \underset{y\in S_x}{lim sup}\frac{f(y)}{g(y)}\le M(x).}$

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

${\displaystyle \underset{x\to c}{lim}m(x)=\underset{x\to c}{lim}M(x)=\underset{x\to c}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=L.}$

and

${\displaystyle \underset{x\to c}{lim}\left(\underset{y\in S_x}{lim inf}\frac{f(y)}{g(y)}\right)=\underset{x\to c}{lim inf}\frac{f(x)}{g(x)}}$ and ${\displaystyle \underset{x\to c}{lim}\left(\underset{y\in S_x}{lim sup}\frac{f(y)}{g(y)}\right)=\underset{x\to c}{lim sup}\frac{f(x)}{g(x)}.}$

In case 1, the squeeze theorem establishes that ${\displaystyle \underset{x\to c}{lim}\frac{f(x)}{g(x)}}$ exists and is equal to L. In the case 2, and the squeeze theorem again asserts that ${\displaystyle \underset{x\to c}{lim inf}\frac{f(x)}{g(x)}=\underset{x\to c}{lim sup}\frac{f(x)}{g(x)}=L}$, and so the limit ${\displaystyle \underset{x\to c}{lim}\frac{f(x)}{g(x)}}$ 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.

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

${\displaystyle f^{\prime }(a)=\underset{x\to a}{lim}{f}^{\prime }(x).}$

In particular, f' is also continuous at a.

Proof

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