Indeterminate Forms and L’Hôpital’s Rule

What is an indeterminate form?

An indeterminate form is a way of summarizing the behavior of a limit when the behavior of the parts does not decide it. Take the four limits \lim_x→0(x^3)/(x)=0 \lim_x→0(x)/(x^3)=∞ \lim_x→0x/x=1 \lim_x→0(2x)/(x)=2. In each, the numerator and the denominator both approach 0, yet the results are 0, ∞, 1 and 2. We say a limit \lim_x→ a(f(x))/(g(x))\quadwith \lim_x→ af(x)=0 and \lim_x→ ag(x)=0 has the indeterminate form \frac00. The symbol "\frac00" looks like a number but is not one: it is not an instruction to divide 0 by 0, only a label for the limit of a ratio of two functions that each go to zero. Knowing that both go to zero is not enough to find the limit of the ratio; one has to look more closely, by algebra or by the rule of this article.

If both the numerator and the denominator go to +∞ or -∞, the form is \frac∞∞. The other forms are ∞·0, ∞-∞, 1^∞, 0^0 and ∞^0. Each is a label for a kind of limit whose value can come out in several ways: x\ln x and x·\frac1x are both of the form 0·∞ at 0^+ and ∞·0 at ∞, and they give 0 and 1.

Some forms are not indeterminate. A numerator that goes to 1 over a denominator that goes to 0 is \frac10, which is infinite (or does not exist if the sign differs on the two sides); \frac0∞ is 0; ∞+∞ is ∞; 0^∞ is 0. Only the forms above leave the limit open.

John (Johann) Bernoulli discovered a rule for using derivatives to calculate limits of fractions whose numerators and denominators both approach zero or ±∞. It is known as l'Hôpital's Rule after Guillaume de l'Hôpital, who wrote the first introductory differential calculus text, where the rule first appeared in print.

What does L’Hôpital’s Rule say?

L'Hôpital's Rule: suppose \lim_x→ af(x)=\lim_x→ ag(x)=0, that f and g are differentiable on an open interval I containing a, and that g'(x)\ne0 on I if x\ne a. Then \lim_x→ a(f(x))/(g(x))=\lim_x→ a(f'(x))/(g'(x)), assuming that the limit on the right side exists. The theorem also applies if x→±∞ or when f(x) and g(x) go to ±∞ (the form \frac∞∞), and to one-sided limits; the number a may be finite or infinite. The reason is that near a a function is close to its tangent line, so the ratio of two functions that vanish at a is the ratio of their slopes. Section 5 proves it.

Caution. To apply the rule to \frac fg, divide the derivative of f by the derivative of g. Do not take the derivative of the quotient: the quotient to use is (f')/(g') and not \left(\frac fg\right)'. For (x^2-1)/(x-1) at 1 the right number is \frac2x1→2, while the derivative of the quotient x+1 is 1.

Using the rule. To find \lim(f(x))/(g(x)), differentiate f and g, and keep differentiating as long as the quotient still has the form \frac00 (or \frac∞∞). As soon as one of the two derivatives no longer goes to zero, stop. The rule does not apply when the numerator or the denominator has a finite limit that is not zero (or, in the first round, when the form is not indeterminate at all).

Example 1. All of these are of the form \frac00, and some need the rule more than once. (a) \lim_x→0(3x-\sin x)/(x)=\lim_x→0(3-\cos x)=2. (b) \lim_x→0(√(1+x)-1)/(x)=\lim_x→0\frac\frac12√(1+x)1=\frac12. (c) \lim_x→0(√(1+x)-1-\frac x2)/(x^2)=\lim_x→0\frac\frac12√(1+x)-\frac122x, which is \frac00 again, so =\lim_x→0(-\frac14(1+x)^-3/2)/(2)=-\frac18. (d) \lim_x→0(x-\sin x)/(x^3)=\lim_x→0(1-\cos x)/(3x^2) =\lim_x→0(\sin x)/(6x)=\lim_x→0(\cos x)/(6)=\frac16.

Example 2: a careful use. \lim_x→0(1-\cos x)/(x+x^2)=\lim_x→0(\sin x)/(1+2x)=\frac01=0. It is tempting to apply the rule again to get \lim(\cos x)/(2)=\frac12, but that is wrong: (\sin x)/(1+2x) is \frac01, not indeterminate, and its value is 0.

Example 3: one-sided limits. (a) \lim_x→0^+(\sin x)/(x^2)=\lim_x→0^+(\cos x)/(2x)=+∞, since \cos x→1 and 2x→0^+. (b) \lim_x→0^-(\sin x)/(x^2)=\lim_x→0^-(\cos x)/(2x)=-∞, since 2x<0. The one-sided limits differ, so the two-sided limit does not exist.

Type a numerator, a denominator and a point below, or choose an example. The stepper applies the rule one round at a time, names the form at each step and stops when the quotient is no longer indeterminate.

How does the rule handle ∞/∞, and which grows faster?

L'Hôpital's Rule applies to \frac∞∞ exactly as to \frac00: if f(x)→±∞ and g(x)→±∞ as x→ a, then \lim(f(x))/(g(x))=\lim(f'(x))/(g'(x)), provided the limit on the right exists or is ±∞. Here a may be finite or infinite, and x→ a may be replaced by x→ a^+ or x→ a^-. The proof of this case is harder and is not given here; it is a 19th-century result.

Example 4. Find the limits of these \frac∞∞ forms. (a) \lim_x→π/2^-(\sec x)/(1+\tan x). Both parts go to +∞ from the left, so =\lim(\sec x\tan x)/(\sec^2x)=\lim\sin x=1 (a simplification, (\tan x)/(\sec x)=\sin x, is easier here than a second round). (b) \lim_x→∞(\ln x)/(2√ x)=\lim_x→∞(1/x)/(1/√ x) =\lim_x→∞\frac1√ x=0. (c) \lim_x→∞(e^x)/(x^2)=\lim(e^x)/(2x)=\lim(e^x)/(2)=∞.

The last two are the comparison of growth rates: a logarithm grows more slowly than any positive power of x, and an exponential grows faster than any power, so (\ln x)/(x^p)→0 and (e^x)/(x^p)→∞ for every p>0. Many limits in later chapters are settled by this ordering: \ln x≪ x^p≪ e^x as x→∞.

Leading terms. For a ratio of polynomials the rule gives the familiar answer: \lim_x→∞(5x^2-3x)/(7x^2+1)=\lim(10x-3)/(14x)=\lim(10)/(14)=\frac57, the ratio of the leading coefficients.

How do you handle 0·∞ and ∞−∞?

The forms ∞·0 and ∞-∞ are not quotients, so first algebra turns them into \frac00 or \frac∞∞, and then the rule applies. Neither ∞·0 nor ∞-∞ is a number; they only describe what the parts do.

Example 5: a product. (a) \lim_x→\inftyx\sin\frac1x is of the form ∞·0. Write it as \lim_x→∞(\sin(1/x))/(1/x), a \frac00 form, and apply the rule: =\lim_x→∞(\cos(1/x)·(-1/x^2))/(-1/x^2) =\lim_x→∞\cos\frac1x=1. (b) \lim_x→0^+x\ln x is of the form 0·∞. Write it as \lim_x→0^+(\ln x)/(1/x), an \frac∞∞ form, and apply the rule: =\lim_x→0^+(1/x)/(-1/x^2)=\lim_x→0^+(-x)=0.

The choice of which factor goes in the denominator matters: x\ln x=(x)/(1/\ln x) would lead to a worse quotient. Put in the denominator the factor whose reciprocal is simpler to differentiate.

Example 6: a difference. Find \lim_x→0\left(\frac1x-\frac1\sin x\right). As x→0^+, \frac1x→∞ and \frac1\sin x→∞, so the form is ∞-∞; as x→0^- it is -∞-(-∞). Neither shows the limit. Combine the fractions over a common denominator, \frac1x-\frac1\sin x=(\sin x-x)/(x\sin x), which is of the form \frac00, and apply the rule twice: \lim_x→0(\sin x-x)/(x\sin x)=\lim_x→0(\cos x-1)/(\sin x+x\cos x)=\lim_x→0(-\sin x)/(2\cos x-x\sin x)=\frac02=0.

Another difference: \lim_x→1^+\left(\frac1\ln x-\frac1x-1\right)=\lim(x-1-\ln x)/((x-1)\ln x)=\frac12, again a combined fraction of the form \frac00 and two rounds.

How do you handle 1^∞, 0^0 and ∞^0?

Limits that lead to 1^∞, 0^0 and ∞^0 are handled by first taking the logarithm of the function, the idea of logarithmic differentiation: find the limit L of \ln f(x) with the rule, then exponentiate. If \lim_x→ a\ln f(x)=L, then \lim_x→ af(x)=\lim_x→ ae^\ln f(x)=e^L, by the continuity of the exponential function. The formula is valid for one-sided limits and for a=±∞. If f(x)=u(x)^v(x) then \ln f=v\ln u, a product, which the previous section turns into the quotient (\ln u)/(1/v).

Example 7. Show that \lim_x→0^+(1+x)^1/x=e. The limit has the form 1^∞. Let f(x)=(1+x)^1/x, so \ln f(x)=(\ln(1+x))/(x), a \frac00 form. By the rule \lim_x→0^+\ln f(x)=\lim_x→0^+(\ln(1+x))/(x)=\lim_x→0^+(1/(1+x))/(1)=1. Therefore \lim_x→0^+(1+x)^1/x=e^1=e. This is the number e from the definition of the exponential, and it is the same as \lim_x→∞\left(1+\frac1x\right)^x=e.

Example 8. Find \lim_x→\inftyx^1/x. The form is ∞^0. With f(x)=x^1/x, \ln f(x)=(\ln x)/(x), an \frac∞∞ form, and \lim_x→∞(\ln x)/(x)=\lim(1/x)/(1)=0. So \lim_x→\inftyx^1/x=e^0=1.

Other examples. \lim_x→0^+x^x has the form 0^0: \ln x^x=x\ln x→0 by Example 5(b), so x^x→ e^0=1. And \lim_x→1^+x^1/(x-1) is 1^∞: (\ln x)/(x-1)→1, so the limit is e. In general \lim_x→∞\left(1+\frac px\right)^x=e^p.

The stepper below takes the form, shows the rewriting and runs the rule for every kind of indeterminate form: 0·∞, ∞-∞ and the powers.

When does L’Hôpital’s Rule not help?

The rule is a tool with conditions, and three things can go wrong: the quotient is not indeterminate, the rule cycles, or the quotient of derivatives has no limit.

1. Not an indeterminate form. If the numerator has a finite limit that is not zero, or the quotient is already determinate, the rule must not be used. \lim_x→2(x^2+1)/(x+1)=\frac53 by substitution; differentiating would give \frac2x1→4, which is wrong. The same happens in the middle of a calculation, as in Example 2, once the quotient has stopped being \frac00.

2. The rule cycles. For \lim_x→∞(√(9x+1))/(√ x+1) the quotient of derivatives is (9√ x)/(√(9x+1)), again of the form \frac∞∞, and the next round brings back a quotient of the same kind: you just keep cycling. Algebra settles it: divide numerator and denominator by √ x to get (√(9+1/x))/(1+1/√ x)→3. Similarly (√(x^2+1))/(x) cycles between a quotient and its reciprocal, and the answer is 1.

3. The derivatives have no limit. The theorem says the limit of \frac fg equals the limit of (f')/(g') if the latter exists. For \lim_x→∞(x+\sin x)/(x) the quotient of derivatives is 1+\cos x, which has no limit, so the rule says nothing, and yet the original limit exists: (x+\sin x)/(x)=1+(\sin x)/(x)→1. In the same way \lim_x→0^+(x^2\sin(1/x))/(\sin x) has derivatives with \cos(1/x), which oscillates, while x\sin\frac1x·(x)/(\sin x)→0.

One-sided limits that differ. Example 3: (\sin x)/(x^2) goes to +∞ from the right and -∞ from the left; the two-sided limit does not exist, and the rule gives the one-sided values correctly.

In all these cases a limit read directly, or a little algebra such as dividing by the dominant term, does what the rule cannot. Try them below: the stepper reports the status, and for a cycle or a missing limit it also reads the original quotient directly.

Why is L’Hôpital’s Rule true?

The rule comes from Cauchy's Mean Value Theorem, an extension of the Mean Value Theorem to two functions.

A picture first. Take f and g with continuous derivatives, f(a)=g(a)=0 and g'(a)\ne0. Near a they are close to their linearizations, f(x)=f'(a)(x-a)+\varepsilon_1(x-a) and g(x)=g'(a)(x-a)+\varepsilon_2(x-a), with \varepsilon_1,\varepsilon_2→0 as x→ a. Then (f(x))/(g(x))=(f'(a)+\varepsilon_1)/(g'(a)+\varepsilon_2)→(f'(a))/(g'(a))=\lim_x→ a(f'(x))/(g'(x)). The ratio of two functions that vanish at a is the ratio of their slopes there. The proof below needs neither g'(a)\ne0 nor continuous derivatives.

Cauchy's Mean Value Theorem. Suppose f and g are continuous on [a,b] and differentiable on (a,b), and g'(x)\ne0 throughout (a,b). Then there is a number c in (a,b) with (f'(c))/(g'(c))=(f(b)-f(a))/(g(b)-g(a)). When g(x)=x this is the Mean Value Theorem.

Proof. First, g(a)\ne g(b): otherwise the Mean Value Theorem would give g'(c)=(g(b)-g(a))/(b-a)=0 for some c, which is excluded. Then apply the Mean Value Theorem to F(x)=f(x)-f(a)-(f(b)-f(a))/(g(b)-g(a)) \big(g(x)-g(a)\big). F is continuous on [a,b] and differentiable on (a,b), and F(a)=F(b)=0. By Rolle's Theorem there is c with F'(c)=0, that is f'(c)-(f(b)-f(a))/(g(b)-g(a)) g'(c)=0, which rearranges to the claim. ∎ Geometrically, the curve (g(t),f(t)) has a tangent parallel to the secant through its ends, and the slope of that tangent is (f')/(g'). Draw a curve below and see the tangents.

Proof of l'Hôpital's Rule (the case x→ a^+, form \frac00). Since f(x)→0 and g(x)→0 and both are differentiable at a, they are continuous there and f(a)=g(a)=0. For x to the right of a apply Cauchy's theorem on [a,x]: there is c between a and x with (f'(c))/(g'(c))=(f(x)-f(a))/(g(x)-g(a))=(f(x))/(g(x)). As x→ a^+, c is squeezed between a and x, so c→ a^+, and therefore \lim_x→ a^+(f(x))/(g(x))=\lim_c→ a^+(f'(c))/(g'(c))=\lim_x→ a^+(f'(x))/(g'(x)). The case x→ a^- is the same on [x,a], and the two together give the two-sided rule. ∎

How do you compute these limits in code?

In code, SymPy's sp.limit(expr, x, a, '+') finds limits exactly, and a few lines of sp.diff reproduce the rule round by round. Without a CAS, a table of the quotient at shrinking distances shows the limit, and the table of (f')/(g') next to it converges to the same number, which is the content of the rule. Numerical evaluation near a point has a trap that the rule does not: for (1+h)^1/h the computer rounds 1+h to 1 once h is below about 10^-16 and returns 1 instead of e, so a table must stop at moderate h.

import sympy as sp

x = sp.symbols('x')


def lhopital(f, g, a, direction='+-', rounds=6):
    """Apply l'Hopital's Rule while the quotient is 0/0 or oo/oo; return the quotient at each round and the limit."""
    steps = []
    for _ in range(rounds):
        lf, lg = sp.limit(f, x, a, direction), sp.limit(g, x, a, direction)
        indeterminate = (lf == 0 and lg == 0) or (lf in (sp.oo, -sp.oo) and lg in (sp.oo, -sp.oo))
        steps.append((sp.simplify(f / g), 'rule' if indeterminate else 'stop'))
        if not indeterminate:
            break
        f, g = sp.diff(f, x), sp.diff(g, x)
    return steps, sp.limit(f / g, x, a, direction)


examples = (
    (3 * x - sp.sin(x), x, 0),                                        # Example 1a
    (x - sp.sin(x), x**3, 0),                                         # Example 1d: three rounds
    (1 - sp.cos(x), x + x**2, 0),                                     # Example 2: one round, then stop
    (sp.exp(x), x**2, sp.oo),                                         # Example 4c
    (sp.sin(x) - x, x * sp.sin(x), 0),                                # Example 6: 1/x - 1/sin x as one fraction
)
for f, g, a in examples:
    print(f, '/', g, 'at', a, '->', lhopital(f, g, a))
print(sp.limit((1 + x) ** (1 / x), x, 0, '+'), sp.limit(x ** (1 / x), x, sp.oo), sp.limit(x ** x, x, 0, '+'))
# 3*x - sin(x) / x at 0 -> ([(3 - sin(x)/x, 'rule'), (3 - cos(x), 'stop')], 2)
# x - sin(x) / x**3 at 0 -> ([((x - sin(x))/x**3, 'rule'), ((1 - cos(x))/(3*x**2), 'rule'), (sin(x)/(6*x), 'rule'), (cos(x)/6, 'stop')], 1/6)
# 1 - cos(x) / x**2 + x at 0 -> ([((1 - cos(x))/(x*(x + 1)), 'rule'), (sin(x)/(2*x + 1), 'stop')], 0)
# exp(x) / x**2 at oo -> ([(exp(x)/x**2, 'rule'), (exp(x)/(2*x), 'rule'), (exp(x)/2, 'stop')], oo)
# -x + sin(x) / x*sin(x) at 0 -> ([((-x/sin(x) + 1)/x, 'rule'), ((cos(x) - 1)/(x*cos(x) + sin(x)), 'rule'), (sin(x)/(x*sin(x) - 2*cos(x)), 'stop')], 0)
# E 1 1

// the quotient f/g at shrinking distances from a: f/g and f'/g' (and the rounds after it) tend to the same number
const near = (f, g, a, hs = [0.1, 0.01, 0.001]) => hs.map((h) => +(f(a + h) / g(a + h)).toFixed(5))
const f = (x) => x - Math.sin(x)
const g = (x) => x ** 3                                              // Example 1d, whose limit is 1/6
near(f, g, 0)                                                        // [0.16658, 0.16667, 0.16667]   f/g
near((x) => 1 - Math.cos(x), (x) => 3 * x * x, 0)                    // [0.16653, 0.16667, 0.16667]   f'/g'
near((x) => Math.sin(x), (x) => 6 * x, 0)                            // [0.16639, 0.16666, 0.16667]   f''/g''
near((x) => Math.cos(x), () => 6, 0)                                 // [0.16583, 0.16666, 0.16667]   f'''/g''': no longer 0/0
// a power limit, and the trap of rounding: (1 + h)^(1/h) tends to e, until 1 + h is rounded to 1
;[1e-2, 1e-4, 1e-6].map((h) => (1 + h) ** (1 / h))                    // [2.7048138294215285, 2.7181459268249255, 2.7182804690957534]
;(1 + 1e-17) ** (1 / 1e-17)                                           // 1: 1 + 1e-17 is 1 in floating point
;[1e-2, 1e-4, 1e-8].map((h) => h * Math.log(h))                       // [-0.046, -0.00092, -1.84e-7]: x ln x -> 0

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

int main() {
  double hs[3];
  hs[0] = 0.1; hs[1] = 0.01; hs[2] = 0.001;
  for (int i = 0; i < 3; i++) {                          // Example 1d: (x - sin x) / x^3, limit 1/6
    double h = hs[i];
    double q0 = (h - sin(h)) / (h * h * h);              // f / g
    double q1 = (1 - cos(h)) / (3 * h * h);              // f' / g'
    double q2 = sin(h) / (6 * h);                        // f'' / g''
    double q3 = cos(h) / 6;                              // f''' / g''', no longer 0/0
    printf("h=%.3f %.5f %.5f %.5f %.5f\n", h, q0, q1, q2, q3);
  }
  printf("%.5f\n", 1.0 / 6);
  printf("%.5f %.5f\n", pow(1.01, 100.0), pow(1.0001, 10000.0));    // (1 + h)^(1/h) tends to e = 2.71828
  return 0;
}

/* output:
h=0.100 0.16658 0.16653 0.16639 0.16583
h=0.010 0.16667 0.16667 0.16666 0.16666
h=0.001 0.16667 0.16667 0.16667 0.16667
0.16667
2.70481 2.71815
*/

Who discovered L’Hôpital’s Rule?

The rule carries the name of the man who published it, not of the man who found it.

• Johann Bernoulli, 1691–92. The Swiss mathematician Johann Bernoulli, a young pupil of the new calculus of Leibniz, taught it in Paris to the French nobleman Guillaume de l'Hôpital, and in the lessons he gave the way of finding the value of a fraction whose numerator and denominator both vanish.
• l'Hôpital, 1696. In his Analyse des infiniment petits pour l'intelligence des lignes courbes, the first textbook of the differential calculus, l'Hôpital printed the rule, saying that he owed much to Bernoulli and Leibniz. Under an agreement l'Hôpital paid Bernoulli for his discoveries and was free to use them.
• A manuscript, 1921. Bernoulli's own lecture notes on the differential calculus were found in the 1920s, and they show that the rule, and much else in the textbook, is Bernoulli's. The name stayed.
• Cauchy, 1823. Augustin-Louis Cauchy gave the general statement and the proof through the mean value theorem for two functions, which carries his name. The original rule dealt with the quotient of two expressions that vanish at a point; repeated use, one-sided limits and the form \frac∞∞ were settled in the 19th century.
• Today. The rule is the first tool for indeterminate limits and the way that limits such as \frac\sin xx and (\ln x)/(x) are taught; computer algebra systems compute them with series, which is faster.

What mistakes do people make with L’Hôpital’s Rule?

The most common mistakes with indeterminate forms and l'Hôpital's Rule are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: L’Hôpital’s Rule at a glance

• Indeterminate forms: \frac00, \frac∞∞, 0·∞, ∞-∞, 1^∞, 0^0, ∞^0 describe limits that the parts do not decide.
• The rule: \frac00 or \frac∞∞ at a gives \lim\frac fg=\lim(f')/(g') if the right side exists or is ±∞; also for one-sided limits and a=±∞.
• How: differentiate f and g separately, repeat while the form stays indeterminate, stop at the first quotient that is not.
• Other forms: products and differences become a quotient by algebra; powers by \ln, then e^L.
• Growth: \ln x≪ x^p≪ e^x as x→∞.
• When it fails: a determinate quotient, a cycle (√(9x+1)/(√ x+1)), or (f')/(g') without a limit ((x+\sin x)/x).
• Proof: Cauchy's Mean Value Theorem, (f'(c))/(g'(c))=(f(b)-f(a))/(g(b)-g(a)), then squeeze c between a and x.
• Next: applied optimization, the closed interval method put to work on real problems.
• Code: sp.limit(expr, x, a, '+'), sp.diff for the rounds, and tables at moderate distances.

Glossary: key terms

Indeterminate form

A label for a limit, such as zero over zero, whose value is not decided by the limits of its parts and can come out as any number or infinity.

L’Hôpital’s Rule

The rule that the limit of a quotient of the form zero over zero or infinity over infinity equals the limit of the quotient of the derivatives, if that limit exists.

Cauchy’s Mean Value Theorem

The Mean Value Theorem for two functions: for some point c, the ratio of the derivatives equals the ratio of the changes of the functions over the interval.

Indeterminate power

A limit of the form one to the infinity, zero to the zero or infinity to the zero, found by taking the logarithm of the function and exponentiating the limit.

One-sided limit

The limit as x approaches a from only the right or only the left; the two-sided limit exists only if both are equal.

Growth rate

How fast a function grows as x goes to infinity; a logarithm grows slower than any power, and an exponential faster than any power.

Frequently asked questions

References

1. Joel Hass, Christopher Heil, Maurice D. Weir et al. (2023). Thomas' Calculus: Early Transcendentals in SI Units (15th ed.), Section 4.5: Indeterminate Forms and L'Hôpital's Rule. Pearson.
2. James Stewart (2015). Calculus (8th ed.), section 4.4: Indeterminate Forms and l'Hospital's Rule. Cengage Learning.
3. Guillaume de l'Hôpital (1696). Analyse des infiniment petits pour l'intelligence des lignes courbes. Paris.
4. Augustin-Louis Cauchy (1823). Résumé des leçons données à l'École royale polytechnique sur le calcul infinitésimal. Paris.

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