L'Hôpital's rule

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Example application of l'Hôpital's rule to Template:Math and Template:Math: the function Template:Math is undefined at Template:Math, but can be completed to a continuous function on all of Template:Math by defining Template:Math.

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Fundamental theorem

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L'Hôpital's rule (Template:IPAc-en, Template:Respell), 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 de 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 Template:Mvar and Template:Mvar which are defined on an open interval Template:Mvar and differentiable on <math display=inline>I\setminus \{c\}</math> for a (possibly infinite) accumulation point Template:Mvar of Template:Mvar, if <math display="inline">\lim \limits_{x\to c}f(x)=\lim \limits_{x\to c}g(x)=0 \text{ or }\pm\infty,</math> and <math display=inline>g'(x)\ne 0</math> for all Template:Mvar in <math display=inline>I\setminus \{c\}</math>, and <math display=inline>\lim \limits_{x\to c}\frac{f'(x)}{g'(x)}</math> exists, then

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

HistoryEdit

Guillaume de l'Hôpital (also written l'HospitalTemplate:Efn) 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.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>Template:Efn However, it is believed that the rule was discovered by the Swiss mathematician Johann Bernoulli.<ref>Template:Cite book Extract of page 321</ref>

General formEdit

The general form of l'Hôpital's rule covers many cases. Let Template:Math and Template:Math be extended real numbers: real numbers, as well as positive and negative infinity. Let Template:Math be an open interval containing Template:Math (for a two-sided limit) or an open interval with endpoint Template:Math (for a one-sided limit, or a limit at infinity if Template:Math is infinite). On <math>I\smallsetminus \{c\}</math>, the real-valued functions Template:Math and Template:Math are assumed differentiable with <math>g'(x) \ne 0</math>. It is also assumed that <math display="inline">\lim \limits_{x\to c} \frac{f'(x)}{g'(x)} = L</math>, a finite or infinite limit.

If either<math display="block">\lim_{x\to c}f(x) = \lim_{x\to c}g(x) = 0</math>or<math display="block">\lim_{x\to c} |f(x)| = \lim_{x\to c} |g(x)| = \infty,</math>then<math display="block">\lim_{x\to c} \frac{f(x)}{g(x)} = L.</math>Although we have written Template:Math throughout, the limits may also be one-sided limits (Template:Math or Template:Math), when Template:Math is a finite endpoint of Template:Math.

In the second case, the hypothesis that Template:Math diverges to infinity is not necessary; in fact, it is sufficient that <math display="inline">\lim_{x \to c} |g(x)| = \infty.</math>

The hypothesis that <math>g'(x)\ne 0</math> appears most commonly in the literature, but some authors sidestep this hypothesis by adding other hypotheses which imply <math>g'(x)\ne 0</math>. For example,<ref>Template:Harv</ref> one may require in the definition of the limit <math display="inline">\lim \limits_{x\to c} \frac{f'(x)}{g'(x)} = L</math> that the function <math display="inline">\frac{f'(x)}{g'(x)}</math> must be defined everywhere on an interval <math>I\smallsetminus \{c\}</math>.Template:Efn Another method<ref>Template:Harv</ref> is to require that both Template:Math and Template:Math be differentiable everywhere on an interval containing Template:Math.

Necessity of conditions: CounterexamplesEdit

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

Indeterminacy of form: <math> \lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0 </math> or <math> \pm \infty </math> ;
Differentiability of functions: <math> f(x) </math> and <math> g(x) </math> are differentiable on an open interval <math> \mathcal{I} </math> except possibly at the limit point <math> c </math> in <math> \mathcal{I} </math>;
Non-zero derivative of denominator: <math> g'(x) \ne 0 </math> for all <math> x </math> in <math> \mathcal{I} </math> with <math> x \ne c </math> ;
Existence of limit of the quotient of the derivatives: <math> \lim_{x \to c} \frac{f'(x)}{g'(x)} </math> exists.

Where one of the above conditions is not satisfied, l'Hôpital's rule is not valid in general, and its conclusion may be false in certain cases.

1. Form is not indeterminateEdit

The necessity of the first condition can be seen by considering the counterexample where the functions are <math> f(x) = x +1 </math> and <math> g(x) = 2x +1 </math> and the limit is <math> x \to 1 </math>.

The first condition is not satisfied for this counterexample because <math> \lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = (1) + 1 = 2 \neq 0 </math> and <math> \lim_{x \to 1} g(x) = \lim_{x \to 1} (2x + 1) = 2(1) + 1 = 3 \neq 0 </math>. This means that the form is not indeterminate.

The second and third conditions are satisfied by <math> f(x) </math> and <math> g(x) </math>. The fourth condition is also satisfied with

But the conclusion fails, since <math display="block"> \lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{x+1}{2x+1} = \frac{ \lim_{x \to 1} (x+1) }{ \lim_{x \to 1} (2x+1) } = \frac{2}{3} \neq \frac{1}{2}. </math>

2. Differentiability of functionsEdit

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

For example, let <math> f(x) = \begin{cases} \sin x, & x\neq0 \\ 1, & x=0 \end{cases} </math> , <math> g(x)=x </math>, and <math> c = 0 </math>. In this case, <math> f(x) </math> is not differentiable at <math> c </math>. However, since <math> f(x) </math> is differentiable everywhere except <math> c </math>, then <math> \lim_{x \to c}f'(x) </math> still exists. Thus, since

<math> \lim_{x\to c} \frac{f(x)}{g(x)} = \frac{0}{0} </math> and <math> \lim_{x\to c} \frac{f'(x)}{g'(x)} </math> exists, l'Hôpital's rule still holds.

3. Derivative of denominator is zeroEdit

The necessity of the condition that <math>g'(x)\ne 0</math> near <math>c</math> can be seen by the following counterexample due to Otto Stolz.<ref name="stolz">Template:Cite journal</ref> Let <math>f(x)=x+\sin x \cos x</math> and <math>g(x)=f(x)e^{\sin x}.</math> Then there is no limit for <math>f(x)/g(x)</math> as <math>x\to\infty.</math> However,

<math>\begin{align}

\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} \\ &= \frac{2\cos x}{2 \cos x +x+\sin x \cos x} e^{-\sin x}, \end{align}</math>

which tends to 0 as <math>x\to\infty</math>, although it is undefined at infinitely many points. Further examples of this type were found by Ralph P. Boas Jr.<ref name="boas">Template:Cite journal</ref>

4. Limit of derivatives does not existEdit

The requirement that the limit <math>\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> exists is essential; if it does not exist, the original limit <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> may nevertheless exist. Indeed, as <math>x</math> approaches <math>c</math>, the functions <math>f</math> or <math>g</math> may exhibit many oscillations of small amplitude but steep slope, which do not affect <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> but do prevent the convergence of <math>\lim_{x\to c}\frac{f'(x)}{g'(x)}</math>.

For example, if <math>f(x)=x+\sin(x)</math>, <math>g(x)=x</math> and <math>c=\infty</math>, then <math display="block">\frac{f'(x)}{g'(x)}=\frac{1+\cos(x)}{1},</math>which does not approach a limit since cosine oscillates infinitely between Template:Math and Template:Math. But the ratio of the original functions does approach a limit, since the amplitude of the oscillations of <math>f</math> becomes small relative to <math>g</math>:

<math>\lim_{x\to\infty}\frac{f(x)}{g(x)}

= \lim_{x\to\infty}\left(\frac{x+\sin(x)}{x}\right) = \lim_{x\to\infty}\left(1+\frac{\sin(x)}{x}\right) = 1+0 = 1. </math>

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

<math> \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)} ,</math>

so that if the limit of <math display="inline">\frac{f}{g} </math> exists, then it must lie between the inferior and superior limits of <math display="inline">\frac{f'}{g'} </math> . In the example, 1 does indeed lie between 0 and 2.)

Note also that by the contrapositive form of the Rule, if <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> does not exist, then <math>\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> also does not exist.

ExamplesEdit

In the following computations, we indicate each application of l'Hôpital's rule by the symbol <math> \ \stackrel{\mathrm{H}}{=}\ </math>.

Here is a basic example involving the exponential function, which involves the indeterminate form Template:Sfrac at Template:Math: <math display="block">

\lim_{x\to 0} \frac{e^x - 1}{x^2+x} \ \stackrel{\mathrm{H}}{=}\  
\lim_{x\to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(x^2+x)} 
 = \lim_{x\to 0} \frac{e^x}{2x+1} = 1.

</math>

This is a more elaborate example involving Template:Sfrac. 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: <math display="block">

\begin{align} \lim_{x\to 0}{\frac{2\sin(x)-\sin(2x)}{x-\sin(x)}} & \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to 0}{\frac{2\cos(x)-2\cos(2x)}{1-\cos(x)}} \\[4pt] & \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to 0}{\frac{-2\sin(x)+4\sin(2x)}{\sin(x)}} \\[4pt] & \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to 0}{\frac{-2\cos(x)+8\cos(2x)}{\cos(x)}}

={\frac{-2+8}{1}} =6.

\end{align}</math>

Here is an example involving Template:Sfrac: <math display="block">

\lim_{x\to\infty}x^n\cdot e^{-x} =\lim_{x\to\infty}{\frac{x^n}{e^x}} \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to\infty}{\frac{nx^{n-1}}{e^x}} =n\cdot \lim_{x\to\infty}{\frac{x^{n-1}}{e^x}}. </math> Repeatedly apply l'Hôpital's rule until the exponent is zero (if Template:Mvar is an integer) or negative (if Template:Mvar is fractional) to conclude that the limit is zero.

Here is an example involving the indeterminate form Template:Math (see below), which is rewritten as the form Template:Sfrac: <math display="block">\lim_{x\to 0^+}x \ln x =\lim_{x\to 0^+} \frac{\ln x}{\frac{1}{x}}

 \ \stackrel{\mathrm{H}}{=}\  \lim_{x\to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}}
 = \lim_{x\to 0^+} -x = 0.</math>

Here is an example involving the mortgage repayment formula and Template:Sfrac. Let Template:Math be the principal (loan amount), Template:Math the interest rate per period and Template:Math the number of periods. When Template:Math is zero, the repayment amount per period is <math>\frac{P}{n}</math> (since only principal is being repaid); this is consistent with the formula for non-zero interest rates: <math display="block"> \lim_{r\to 0}\frac{Pr(1+r)^n}{(1+r)^n-1}

   \ \stackrel{\mathrm{H}}{=}\  P \lim_{r\to 0} \frac{(1+r)^n+rn(1+r)^{n-1}}{n(1+r)^{n-1}}
   = \frac{P}{n}.</math>

One can also use l'Hôpital's rule to prove the following theorem. If Template:Math is twice-differentiable in a neighborhood of Template:Math and its second derivative is continuous on this neighborhood, then <math display="block">\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} \\[4pt]
   &= \lim_{h\to 0}\frac{f(x+h) + f(x-h)}{2} \\[4pt]
   &= f(x).
 \end{align}</math>

Sometimes l'Hôpital's rule is invoked in a tricky way: suppose <math> f(x) + f'(x)</math> converges as Template:Math and that <math>e^x\cdot f(x)</math> converges to positive or negative infinity. Then:<math display="block">

\lim_{x\to\infty }f(x) = \lim_{x\to\infty}\frac{e^x\cdot f(x)}{e^x} \ \stackrel{\mathrm{H}}{=}\ \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),

</math>and so, <math display="inline">\lim_{x\to\infty}f(x)</math> exists and <math display="inline">\lim_{x\to\infty}f'(x) = 0.</math> (This result remains true without the added hypothesis that <math>e^x\cdot f(x)</math> converges to positive or negative infinity, but the justification is then incomplete.)

ComplicationsEdit

Sometimes L'Hôpital's rule does not reduce to an obvious limit in a finite number of steps, unless some intermediate simplifications are applied. Examples include the following:

Two applications can lead to a return to the original expression that was to be evaluated: <math display="block">

\lim_{x\to\infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to\infty} \frac{e^x-e^{-x}}{e^x+e^{-x}} \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to\infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} \ \stackrel{\mathrm{H}}{=}\ \cdots . </math> This situation can be dealt with by substituting <math>y=e^x</math> and noting that Template:Math goes to infinity as Template:Math goes to infinity; with this substitution, this problem can be solved with a single application of the rule: <math display="block"> \lim_{x\to\infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} = \lim_{y\to\infty} \frac{y+y^{-1}}{y-y^{-1}} \ \stackrel{\mathrm{H}}{=}\ \lim_{y\to\infty} \frac{1-y^{-2}}{1+y^{-2}} = \frac{1}{1} = 1. </math> Alternatively, the numerator and denominator can both be multiplied by <math>e^x,</math> at which point L'Hôpital's rule can immediately be applied successfully:<ref>Multiplying by <math>e^{-x}</math> instead yields a solution to the limit without need for l'Hôpital's rule.</ref> <math display="block"> \lim_{x\to\infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} = \lim_{x\to\infty} \frac{e^{2x} + 1}{e^{2x} - 1} \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to\infty} \frac{2e^{2x}}{2e^{2x}} = 1.</math>

An arbitrarily large number of applications may never lead to an answer even without repeating:<math display="block">

\lim_{x\to\infty} \frac{x^\frac1{2}+x^{-\frac1{2}}}{x^\frac1{2}-x^{-\frac1{2}}} \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to\infty} \frac{\frac1{2}x^{-\frac1{2}}-\frac{1}{2}x^{-\frac3{2}}}{\frac1{2}x^{-\frac1{2}}+\frac1{2}x^{-\frac3{2}}} \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to\infty} \frac{-\frac1{4}x^{-\frac3{2}}+\frac3{4}x^{-\frac5{2}}}{-\frac1{4}x^{-\frac3{2}}-\frac3{4}x^{-\frac5{2}}} \ \stackrel{\mathrm{H}}{=}\ \cdots .</math>This situation too can be dealt with by a transformation of variables, in this case <math>y = \sqrt{x}</math>: <math display="block"> \lim_{x\to\infty} \frac{x^\frac1{2}+x^{-\frac1{2}}}{x^\frac1{2}-x^{-\frac1{2}}} = \lim_{y\to\infty} \frac{y+y^{-1}}{y-y^{-1}} \ \stackrel{\mathrm{H}}{=}\ \lim_{y\to\infty} \frac{1-y^{-2}}{1+y^{-2}} = \frac1{1} = 1. </math> Again, an alternative approach is to multiply numerator and denominator by <math>x^{1/2}</math> before applying L'Hôpital's rule: <math display="block"> \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} \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to\infty} \frac{1}{1} = 1.</math>

A common logical fallacy is to use L'Hôpital's rule to prove the value of a derivative by computing the limit of a difference quotient. Since applying l'Hôpital requires knowing the relevant derivatives, this amounts to circular reasoning or begging the question, assuming what is to be proved. For example, consider the proof of the derivative formula for powers of x:

Applying L'Hôpital's rule and finding the derivatives with respect to Template:Math yields Template:Math as expected, but this computation requires the use of the very formula that is being proven. Similarly, to prove <math>\lim_{x\to 0}\frac{\sin(x)}{x}=1</math>, applying L'Hôpital requires knowing the derivative of <math>\sin(x)</math> at <math>x=0</math>, which amounts to calculating <math>\lim_{h\to 0}\frac{\sin(h)}{h}</math> in the first place; a valid proof requires a different method such as the squeeze theorem.

Other indeterminate formsEdit

Other indeterminate forms, such as Template:Math, Template:Math, Template:Math, Template:Math, and Template:Math, can sometimes be evaluated using L'Hôpital's rule. We again indicate applications of L'Hopital's rule by <math> \ \stackrel{\mathrm{H}}{=}\ </math>.

For example, to evaluate a limit involving Template:Math, convert the difference of two functions to a quotient:

<math>

\begin{align} \lim_{x\to 1}\left(\frac{x}{x-1}-\frac1{\ln x}\right) & = \lim_{x\to 1}\frac{x\cdot\ln x -x+1}{(x-1)\cdot\ln x} \\[6pt] & \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to 1}\frac{\ln x}{\frac{x-1}{x}+\ln x} \\[6pt] & = \lim_{x\to 1}\frac{x\cdot\ln x}{x-1+x\cdot\ln x} \\[6pt] & \ \stackrel{\mathrm{H}}{=}\ \lim_{x\to 1}\frac{1+\ln x}{1+1+\ln x} = \frac{1+0}{1+1+0}. \end{align} </math>

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 Template:Math:

<math>

\lim_{x\to 0^+\!}x^x = \lim_{x\to 0^+\!}e^{\ln (x^x)} = \lim_{x\to 0^+\!}e^{x\cdot\ln x} = \lim_{x\to 0^+\!}\exp(x\cdot\ln x) = \exp({\lim\limits_{x\to 0^+\!\!}\,x\cdot\ln x}). </math>

It is valid to move the limit inside the exponential function because this function is continuous. Now the exponent <math>x</math> has been "moved down". The limit <math>\lim_{x\to 0^+}x\cdot\ln x</math> is of the indeterminate form Template:Math dealt with in an example above: L'Hôpital may be used to determine that

Thus

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

Indeterminate form with f & g	Conditions	Transformation to <math>0/0</math>
Template:Sfrac	<math> \lim_{x \to c} f(x) = 0,\ \lim_{x \to c} g(x) = 0 \! </math>	check\|unknown=\|preview=Page using Template:Center with unknown parameter "_VALUE_"\|ignoreblank=y\| 1 \| style }}
Template:Sfrac	<math> \lim_{x \to c} f(x) = \infty,\ \lim_{x \to c} g(x) = \infty \! </math>	<math> \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{1/g(x)}{1/f(x)} \! </math>
<math>0\cdot\infty</math>	<math> \lim_{x \to c} f(x) = 0,\ \lim_{x \to c} g(x) = \infty \! </math>	<math> \lim_{x \to c} f(x)g(x) = \lim_{x \to c} \frac{f(x)}{1/g(x)} \! </math>
<math>\infty - \infty</math>	<math> \lim_{x \to c} f(x) = \infty,\ \lim_{x \to c} g(x) = \infty \! </math>	<math> \lim_{x \to c} (f(x) - g(x)) = \lim_{x \to c} \frac{1/g(x) - 1/f(x)}{1/(f(x)g(x))} \! </math>
<math>0^0</math>	<math> \lim_{x \to c} f(x) = 0^+, \lim_{x \to c} g(x) = 0 \! </math>	<math> \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{g(x)}{1/\ln f(x)} \! </math>
<math>1^\infty</math>	<math> \lim_{x \to c} f(x) = 1,\ \lim_{x \to c} g(x) = \infty \! </math>	<math> \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} \! </math>
<math>\infty^0</math>	<math> \lim_{x \to c} f(x) = \infty,\ \lim_{x \to c} g(x) = 0 \! </math>	<math> \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{g(x)}{1/\ln f(x)} \! </math>

Stolz–Cesàro theoremEdit

Template:Main article

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

Geometric interpretation: parametric curve and velocity vectorEdit

Consider the parametric curve in the xy-plane with coordinates given by the continuous functions <math>g(t)</math> and <math>f(t)</math>, the locus of points <math>(g(t),f(t))</math>, and suppose <math>f(c) = g(c) = 0</math>. The slope of the tangent to the curve at <math>(g(c),f(c))=(0,0)</math> is the limit of the ratio <math>\textstyle \frac{f(t)}{g(t)}</math> as Template:Math. The tangent to the curve at the point <math>(g(t),f(t))</math> is the velocity vector <math>(g'(t),f'(t))</math> with slope <math>\textstyle \frac{f'(t)}{g'(t)}</math>. L'Hôpital's rule then states that the slope of the curve at the origin (Template:Math) is the limit of the tangent slope at points approaching the origin, provided that this is defined.

Proof of L'Hôpital's ruleEdit

Special caseEdit

The proof of L'Hôpital's rule is simple in the case where Template:Math and Template:Math are continuously differentiable at the point Template:Math and where a finite limit is found after the first round of differentiation. This is only a special case of L'Hôpital's rule, because it only applies to functions satisfying stronger conditions than required by the general rule. However, many common functions have continuous derivatives (e.g. polynomials, sine and cosine, exponential functions), so this special case covers most applications.

Suppose that Template:Math and Template:Math are continuously differentiable at a real number Template:Math, that <math>f(c)=g(c)=0</math>, and that <math>g'(c)\neq 0</math>. Then

<math>

\begin{align} & \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{align} </math>

This follows from the difference quotient definition of the derivative. The last equality follows from the continuity of the derivatives at Template:Math. The limit in the conclusion is not indeterminate because <math>g'(c)\ne 0</math>.

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

General proofEdit

The following proof is due to Template:Harvtxt, where a unified proof for the <math display="inline">\frac{0}{0}</math> and <math display="inline">\frac{\pm \infty}{\pm \infty} </math> indeterminate forms is given. Taylor notes that different proofs may be found in Template:Harvtxt and Template:Harvtxt.

Let f and g be functions satisfying the hypotheses in the General form section. Let <math>\mathcal{I}</math> be the open interval in the hypothesis with endpoint c. Considering that <math>g'(x)\ne 0</math> on this interval and g is continuous, <math>\mathcal{I}</math> can be chosen smaller so that g is nonzero on <math>\mathcal{I}</math>.Template:Efn

For each x in the interval, define <math>m(x)=\inf\frac{f'(t)}{g'(t)}</math> and <math>M(x)=\sup\frac{f'(t)}{g'(t)}</math> as <math>t</math> 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 <math>\mathcal{I}</math>, Cauchy's mean value theorem ensures that for any two distinct points x and y in <math>\mathcal{I}</math> there exists a <math>\xi</math> between x and y such that <math>\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(\xi)}{g'(\xi)}</math>. Consequently, <math>m(x)\leq \frac{f(x)-f(y)}{g(x)-g(y)} \leq M(x)</math> 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 Template:Sfrac.

Case 1: <math>\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0</math>

For any x in the interval <math>\mathcal{I}</math>, and point y between x and c,

and therefore as y approaches c, <math>\frac{f(y)}{g(x)}</math> and <math>\frac{g(y)}{g(x)}</math> become zero, and so

Case 2: <math>\lim_{x\to c}|g(x)|=\infty</math>

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

As y approaches c, both <math>\frac{f(x)}{g(y)}</math> and <math>\frac{g(x)}{g(y)}</math> become zero, and therefore

<math>m(x)\le \liminf_{y\in S_x} \frac{f(y)}{g(y)} \le \limsup_{y\in S_x} \frac{f(y)}{g(y)} \le M(x).</math>

The limit superior and limit inferior are necessary since the existence of the limit of Template:Sfrac has not yet been established.

It is also the case that

Template:Efn and

<math>\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)}</math> and <math>\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)}. </math>

In case 1, the squeeze theorem establishes that <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> exists and is equal to L. In the case 2, and the squeeze theorem again asserts that <math>\liminf_{x\to c}\frac{f(x)}{g(x)}=\limsup_{x\to c}\frac{f(x)}{g(x)}=L</math>, and so the limit <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> 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.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

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.

CorollaryEdit

A simple but very useful consequence of L'Hopital's rule is that the derivative of a function cannot have a removable discontinuity. That is, suppose that f is continuous at a, and that <math>f'(x)</math> exists for all x in some open interval containing a, except perhaps for <math>x = a</math>. Suppose, moreover, that <math>\lim_{x\to a}f'(x)</math> exists. Then <math>f'(a)</math> also exists and

In particular, f' is also continuous at a.

Thus, if a function is not continuously differentiable near a point, the derivative must have an essential discontinuity at that point.

ProofEdit

Consider the functions <math>h(x) = f(x)-f(a)</math> and <math>g(x) = x-a</math>. The continuity of f at a tells us that <math>\lim_{x\to a}h(x) = 0</math>. Moreover, <math>\lim_{x\to a}g(x) = 0</math> since a polynomial function is always continuous everywhere. Applying L'Hopital's rule shows that <math>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)</math>.

NotesEdit

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ReferencesEdit

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SourcesEdit

Template:Calculus topics

L'Hôpital's rule

Contents

HistoryEdit

General formEdit

Necessity of conditions: CounterexamplesEdit

1. Form is not indeterminateEdit

2. Differentiability of functionsEdit

3. Derivative of denominator is zeroEdit

4. Limit of derivatives does not existEdit

ExamplesEdit

ComplicationsEdit

Other indeterminate formsEdit

Stolz–Cesàro theoremEdit

Geometric interpretation: parametric curve and velocity vectorEdit

Proof of L'Hôpital's ruleEdit

Special caseEdit

General proofEdit

CorollaryEdit

ProofEdit

See alsoEdit

NotesEdit

ReferencesEdit

SourcesEdit