L'Hôpital's rule

L'Hôpital's rule (/ˌloʊpiːˈtɑːl/, loh-pee-TAHL) or L'Hospital's rule, 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.

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 defined on an open interval I and differentiable on ${\textstyle I\setminus \{c\}}$ for a (possibly infinite) accumulation point c of I, if ${\textstyle \lim \limits _{x\to c}f(x)=\lim \limits _{x\to c}g(x)=0{\text{ or }}\pm \infty ,}$ and ${\textstyle g'(x)\neq 0}$ for all x in I with x ≠ c, and ${\textstyle \lim \limits _{x\to c}{\frac {f'(x)}{g'(x)}}}$ 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 directly evaluated by continuity.

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]

The general form of L'Hôpital's rule covers many cases. Let c and L be extended real numbers: real numbers, positive 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). On ${\displaystyle I\smallsetminus \{c\}}$, the real-valued functions f and g are assumed differentiable with ${\displaystyle g'(x)\neq 0}$. It is also assumed that ${\textstyle \lim \limits _{x\to c}{\frac {f'(x)}{g'(x)}}=L}$, a finite or infinite limit.

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 necessary; in fact, it is sufficient that ${\textstyle \lim _{x\to c}|g(x)|=\infty .}$

The hypothesis that ${\displaystyle g'(x)\neq 0}$ appears most commonly in the literature, but some authors sidestep this hypothesis by adding other hypotheses which imply ${\displaystyle g'(x)\neq 0}$. For example,^[4] one may require in the definition of the limit ${\textstyle \lim \limits _{x\to c}{\frac {f'(x)}{g'(x)}}=L}$ that the function ${\textstyle {\frac {f'(x)}{g'(x)}}}$ must be defined everywhere on an interval ${\displaystyle I\smallsetminus \{c\}}$.^{[lower-alpha 3]} Another method^[5] is to require that both f and g be differentiable everywhere on an interval containing c.

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

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

Where one of the above conditions is not satisfied, the conclusion of L'Hôpital's rule will be false in certain cases.

1. 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 ${\displaystyle \lim _{x\to 1}f(x)=\lim _{x\to 1}(x+1)=(1)+1=2\neq 0}$ and ${\displaystyle \lim _{x\to 1}g(x)=\lim _{x\to 1}(2x+1)=2(1)+1=3\neq 0}$. This means that the form is not indeterminate.

The second and third conditions are satisfied by ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$. The fourth condition is also satisfied with

$${\displaystyle \lim _{x\to 1}{\frac {f'(x)}{g'(x)}}=\lim _{x\to 1}{\frac {(x+1)'}{(2x+1)'}}=\lim _{x\to 1}{\frac {1}{2}}={\frac {1}{2}}.}$$

But the conclusion fails, since $${\displaystyle \lim _{x\to 1}{\frac {f(x)}{g(x)}}=\lim _{x\to 1}{\frac {x+1}{2x+1}}={\frac {\lim _{x\to 1}(x+1)}{\lim _{x\to 1}(2x+1)}}={\frac {2}{3}}\neq {\frac {1}{2}}.}$$

2. 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 ${\displaystyle {\mathcal {I}}}$. The fact that ${\displaystyle {\mathcal {I}}}$ 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 ${\displaystyle c}$ exists because L'Hôpital's rule only requires the derivative to exist as the function approaches ${\displaystyle c}$; the derivative does not need to be taken at ${\displaystyle c}$.

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

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

3. Derivative of 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.^[6] 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,

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

which tends to 0 as ${\displaystyle x\to \infty }$, although it is undefined at infinitely many points. Further examples of this type were found by Ralph P. Boas Jr.^[7]

4. Limit of derivatives does not exist

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

For example, if ${\displaystyle f(x)=x+\sin(x)}$, ${\displaystyle g(x)=x}$ and ${\displaystyle c=\infty }$, then $${\displaystyle {\frac {f'(x)}{g'(x)}}={\frac {1+\cos(x)}{1}},}$$which does not approach a limit since cosine oscillates infinitely between 1 and −1. But the ratio of the original functions does approach a limt, since the amplitude of the oscillations of ${\displaystyle f}$ becomes small relative to ${\displaystyle g}$:

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

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 ${\textstyle {\frac {f}{g}}}$ exists, then it must lie between the inferior and superior limits of ${\textstyle {\frac {f'}{g'}}}$ . In the example, 1 does indeed lie between 0 and 2.)

Note also that by the contrapositive form of the Rule, if ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}}$ does not exist, then ${\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$ also does not exist.

In the following computations, we indicate each application of L'Hopital's rule by the symbol ${\displaystyle \ {\stackrel {\mathrm {H} }{=}}\ }$.

• Here is a basic example involving the exponential function, which involves the indeterminate form ⁠0/0⁠ at x = 0: $${\displaystyle \lim _{x\to 0}{\frac {e^{x}-1}{x^{2}+x}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to 0}{\frac {{\frac {d}{dx}}(e^{x}-1)}{{\frac {d}{dx}}(x^{2}+x)}}=\lim _{x\to 0}{\frac {e^{x}}{2x+1}}=1.}$$
• 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: $${\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {2\sin(x)-\sin(2x)}{x-\sin(x)}}&\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to 0}{\frac {2\cos(x)-2\cos(2x)}{1-\cos(x)}}\\[4pt]&\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to 0}{\frac {-2\sin(x)+4\sin(2x)}{\sin(x)}}\\[4pt]&\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to 0}{\frac {-2\cos(x)+8\cos(2x)}{\cos(x)}}={\frac {-2+8}{1}}=6.\end{aligned}}}$$
• Here is an example involving ⁠∞/∞⁠: $${\displaystyle \lim _{x\to \infty }x^{n}\cdot e^{-x}=\lim _{x\to \infty }{\frac {x^{n}}{e^{x}}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to \infty }{\frac {nx^{n-1}}{e^{x}}}=n\cdot \lim _{x\to \infty }{\frac {x^{n-1}}{e^{x}}}.}$$ 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 ⁠∞/∞⁠: $${\displaystyle \lim _{x\to 0^{+}}x\ln x=\lim _{x\to 0^{+}}{\frac {\ln x}{\frac {1}{x}}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to 0^{+}}{\frac {\frac {1}{x}}{-{\frac {1}{x^{2}}}}}=\lim _{x\to 0^{+}}-x=0.}$$
• 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: $${\displaystyle \lim _{r\to 0}{\frac {Pr(1+r)^{n}}{(1+r)^{n}-1}}\ {\stackrel {\mathrm {H} }{=}}\ P\lim _{r\to 0}{\frac {(1+r)^{n}+rn(1+r)^{n-1}}{n(1+r)^{n-1}}}={\frac {P}{n}}.}$$
• One can also use L'Hôpital's rule to prove the following theorem. If f is twice-differentiable in a neighborhood of x and its second derivative is continuous on this neighborhood, 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 ${\displaystyle f(x)+f'(x)}$ converges as x → ∞ and that ${\displaystyle e^{x}\cdot f(x)}$ converges to positive or negative infinity. Then:$${\displaystyle \lim _{x\to \infty }f(x)=\lim _{x\to \infty }{\frac {e^{x}\cdot f(x)}{e^{x}}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to \infty }{\frac {e^{x}{\bigl (}f(x)+f'(x){\bigr )}}{e^{x}}}=\lim _{x\to \infty }{\bigl (}f(x)+f'(x){\bigr )},}$$and so, ${\textstyle \lim _{x\to \infty }f(x)}$ exists and ${\textstyle \lim _{x\to \infty }f'(x)=0.}$ (This 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.)

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: $${\displaystyle \lim _{x\to \infty }{\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to \infty }{\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to \infty }{\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}\ {\stackrel {\mathrm {H} }{=}}\ \cdots .}$$ 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: $${\displaystyle \lim _{x\to \infty }{\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}=\lim _{y\to \infty }{\frac {y+y^{-1}}{y-y^{-1}}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{y\to \infty }{\frac {1-y^{-2}}{1+y^{-2}}}={\frac {1}{1}}=1.}$$ 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:^[8] $${\displaystyle \lim _{x\to \infty }{\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}=\lim _{x\to \infty }{\frac {e^{2x}+1}{e^{2x}-1}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to \infty }{\frac {2e^{2x}}{2e^{2x}}}=1.}$$
• An arbitrarily large number of applications may never lead to an answer even without repeating:$${\displaystyle \lim _{x\to \infty }{\frac {x^{\frac {1}{2}}+x^{-{\frac {1}{2}}}}{x^{\frac {1}{2}}-x^{-{\frac {1}{2}}}}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to \infty }{\frac {{\frac {1}{2}}x^{-{\frac {1}{2}}}-{\frac {1}{2}}x^{-{\frac {3}{2}}}}{{\frac {1}{2}}x^{-{\frac {1}{2}}}+{\frac {1}{2}}x^{-{\frac {3}{2}}}}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to \infty }{\frac {-{\frac {1}{4}}x^{-{\frac {3}{2}}}+{\frac {3}{4}}x^{-{\frac {5}{2}}}}{-{\frac {1}{4}}x^{-{\frac {3}{2}}}-{\frac {3}{4}}x^{-{\frac {5}{2}}}}}\ {\stackrel {\mathrm {H} }{=}}\ \cdots .}$$This situation too can be dealt with by a transformation of variables, in this case ${\displaystyle y={\sqrt {x}}}$: $${\displaystyle \lim _{x\to \infty }{\frac {x^{\frac {1}{2}}+x^{-{\frac {1}{2}}}}{x^{\frac {1}{2}}-x^{-{\frac {1}{2}}}}}=\lim _{y\to \infty }{\frac {y+y^{-1}}{y-y^{-1}}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{y\to \infty }{\frac {1-y^{-2}}{1+y^{-2}}}={\frac {1}{1}}=1.}$$ Again, an alternative approach is to multiply numerator and denominator by ${\displaystyle x^{1/2}}$ before applying L'Hôpital's rule: $${\displaystyle \lim _{x\to \infty }{\frac {x^{\frac {1}{2}}+x^{-{\frac {1}{2}}}}{x^{\frac {1}{2}}-x^{-{\frac {1}{2}}}}}=\lim _{x\to \infty }{\frac {x+1}{x-1}}\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to \infty }{\frac {1}{1}}=1.}$$

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:

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

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

Other indeterminate forms, such as 1^∞, 0⁰, ∞⁰, 0 · ∞, and ∞ − ∞, can sometimes be evaluated using L'Hôpital's rule. We again indicate applications of L'Hopital's rule by ${\displaystyle \ {\stackrel {\mathrm {H} }{=}}\ }$.

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

${\displaystyle {\begin{aligned}\lim _{x\to 1}\left({\frac {x}{x-1}}-{\frac {1}{\ln x}}\right)&=\lim _{x\to 1}{\frac {x\cdot \ln x-x+1}{(x-1)\cdot \ln x}}\\[6pt]&\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to 1}{\frac {\ln x}{{\frac {x-1}{x}}+\ln x}}\\[6pt]&=\lim _{x\to 1}{\frac {x\cdot \ln x}{x-1+x\cdot \ln x}}\\[6pt]&\ {\stackrel {\mathrm {H} }{=}}\ \lim _{x\to 1}{\frac {1+\ln x}{1+1+\ln x}}={\frac {1+0}{1+1+0}}.\end{aligned}}}$

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:

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

It is valid to move the limit inside the exponential function because this 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 · ∞ dealt with in an example above: L'Hôpital may be used to determine that

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

Thus

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

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 ${\displaystyle 0/0}$
⁠0/0⁠	${\displaystyle \lim _{x\to c}f(x)=0,\ \lim _{x\to c}g(x)=0\!}$	—
⁠${\displaystyle \infty }$/${\displaystyle \infty }$⁠	${\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=\infty \!}$	${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {1/g(x)}{1/f(x)}}\!}$
${\displaystyle 0\cdot \infty }$	${\displaystyle \lim _{x\to c}f(x)=0,\ \lim _{x\to c}g(x)=\infty \!}$	${\displaystyle \lim _{x\to c}f(x)g(x)=\lim _{x\to c}{\frac {f(x)}{1/g(x)}}\!}$
${\displaystyle \infty -\infty }$	${\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=\infty \!}$	${\displaystyle \lim _{x\to c}(f(x)-g(x))=\lim _{x\to c}{\frac {1/g(x)-1/f(x)}{1/(f(x)g(x))}}\!}$
${\displaystyle 0^{0}}$