L'Hôpital's rule

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]

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 ${\textstyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}=L.}$ 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 ${\textstyle \lim _{x\to c}|g(x)|=\infty .}$

The hypothesis that ${\displaystyle g'(x)\neq 0}$ appears most commonly in the literature, but some authors sidestep this hypothesis by adding other hypotheses 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.

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

1. Indeterminancy of form: ${\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0}$ or ${\displaystyle \pm \infty }$ ; and
2. Differentiability of functions: ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$ are differentiable on an open interval ${\displaystyle {\mathcal {I}}}$ except possibly at a point ${\displaystyle c}$ contained in ${\displaystyle {\mathcal {I}}}$ (the same point from the limit) ; and
3. Non-zero derivative of denominator: ${\displaystyle g'(x)\neq 0}$ for all ${\displaystyle x}$ in ${\displaystyle {\mathcal {I}}}$ with ${\displaystyle x\neq c}$ ; and
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, L'Hôpital's rule is not valid in general, and so it cannot always be applied.

• 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 and that its second derivative is continuous on this neighbourhood, 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, ${\textstyle \lim _{x\to \infty }f(x)}$ exists and ${\textstyle \lim _{x\to \infty }f'(x)=0.}$

  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.

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:[9]
  $${\displaystyle \lim _{x\to \infty }{\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}=\lim _{x\to \infty }{\frac {e^{2x}+1}{e^{2x}-1}}=\lim _{x\to \infty }{\frac {2e^{2x}}{2e^{2x}}}=1.}$$

  
• 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 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.

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

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

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

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

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

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.

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.

• L'Hôpital controversy