L'Hôpital's Rule: Resolving "0/0" and Its Relatives

1. What is an indeterminate form?

Many limits can be computed by simply plugging in. But when both \( f(x) \to 0 \) and \( \varphi(x) \to 0 \), the quotient \( \frac{f(x)}{\varphi(x)} \) can tend to anything:

\[ \frac{x^2}{x} \to 0, \qquad \frac{x}{x^2} \to \infty, \qquad \frac{5x}{x} \to 5 \qquad (x \to 0). \]

All three are of the "form \( \frac00 \)", yet their limits differ. The form alone doesn't determine the answer; that's why it is called indeterminate. What decides the limit is the race between numerator and denominator: which one goes to zero faster.

There are seven indeterminate forms:

\[ \frac00, \qquad \frac\infty\infty, \qquad 0\cdot\infty, \qquad \infty - \infty, \qquad 1^\infty, \qquad 0^0, \qquad \infty^0. \]

L'Hôpital's rule handles the first two directly. The other five are reduced to them by algebra.

2. The rule for \(\frac00\) and \(\frac\infty\infty\)

L'Hôpital's rule If the functions \(f(x)\) and \(\varphi(x)\) are both infinitely small or both infinitely large as \(x \to a\), that is, if \( \frac{f(x)}{\varphi(x)} \) is an indeterminate expression of the form \( \frac00 \) or \( \frac\infty\infty \) as \(x \to a\), then \[ \lim_{x\to a}\frac{f(x)}{\varphi(x)} = \lim_{x\to a}\frac{f'(x)}{\varphi'(x)}, \] provided the limit on the right exists. The rule also holds when \(a = \infty\).

The precise conditions: \(f\) and \(\varphi\) are differentiable near \(a\) (not necessarily at \(a\) itself), and \( \varphi'(x) \ne 0 \) near \(a\). The limit on the right may be finite or infinite.

The key point: differentiate the numerator and the denominator separately. This is not the quotient rule.

If \( \frac{f'}{\varphi'} \) is again of the form \( \frac00 \) or \( \frac\infty\infty \), apply the rule again, as many times as needed (as long as the conditions keep holding).

3. Why it works

Take the case \( \frac00 \) with \( f(a) = \varphi(a) = 0 \). By Cauchy's mean value theorem (previous post), for \(x\) near \(a\):

\[ \frac{f(x)}{\varphi(x)} = \frac{f(x) - f(a)}{\varphi(x) - \varphi(a)} = \frac{f'(\xi)}{\varphi'(\xi)}, \qquad \xi \text{ between } a \text{ and } x. \]

As \( x \to a \), the point \(\xi\), squeezed between \(a\) and \(x\), also tends to \(a\). So if \( \frac{f'}{\varphi'} \) has a limit, \( \frac{f}{\varphi} \) has the same limit. That's the whole proof.

Intuition: near \(a\), both functions are almost linear (the differential): \( f(x) \approx f'(a)(x - a) \), \( \varphi(x) \approx \varphi'(a)(x - a) \). The factor \( (x - a) \) cancels, leaving the ratio of slopes. When two functions both reach zero, what matters is how fast each one is moving when it gets there.

The case \(a = \infty\) is reduced to \(a = 0\) by the substitution \( x = \frac1t \); the case \( \frac\infty\infty \) needs a slightly more careful argument, but the conclusion is the same.

4. Worked examples

Example 1 · The fundamental trigonometric limit \[ \lim_{x\to 0}\frac{\sin x}{x} \;\left[\tfrac00\right] = \lim_{x\to 0}\frac{\cos x}{1} = 1. \] (Strictly speaking this is circular, because the derivative of \(\sin x\) was computed using this very limit. Use it as a check, not a proof.)

Example 2 · Applying the rule twice \[ \lim_{x\to 0}\frac{1 - \cos x}{x^2} \;\left[\tfrac00\right] = \lim_{x\to 0}\frac{\sin x}{2x} \;\left[\tfrac00\right] = \lim_{x\to 0}\frac{\cos x}{2} = \frac12. \]

Example 3 · Three times \[ \lim_{x\to 0}\frac{x - \sin x}{x^3} = \lim_{x\to 0}\frac{1 - \cos x}{3x^2} = \lim_{x\to 0}\frac{\sin x}{6x} = \frac16. \] This agrees with the Taylor series method of the post on power series, where \( x - \sin x = \frac{x^3}{6} - \cdots \).

Example 4 · The form \(\frac\infty\infty\) \[ \lim_{x\to\infty}\frac{\ln x}{x} \;\left[\tfrac\infty\infty\right] = \lim_{x\to\infty}\frac{1/x}{1} = 0. \] The logarithm grows, but much more slowly than \(x\).

Example 5 · Repeated application For any natural \(n\), \[ \lim_{x\to\infty}\frac{x^n}{e^x} = \lim_{x\to\infty}\frac{nx^{n-1}}{e^x} = \cdots = \lim_{x\to\infty}\frac{n!}{e^x} = 0. \] After \(n\) steps the power has disappeared, while \(e^x\) is unchanged by differentiation. The exponential beats every power.

Example 6 · Simplify between the steps \[ \lim_{x\to 0}\frac{e^x - e^{-x}}{\sin x} = \lim_{x\to 0}\frac{e^x + e^{-x}}{\cos x} = \frac{2}{1} = 2. \] In longer problems, cancel common factors and replace factors with known limits (like \( \frac{\sin x}{x} \to 1 \)) between applications. Otherwise the derivatives can grow messier at each step.

5. Common mistakes and limitations

Mistake 1 · Using the rule when the form is not indeterminate \[ \lim_{x\to 1}\frac{x^2 + 1}{x + 1} = \frac22 = 1. \] Applying the rule anyway would give \( \lim_{x\to1}\frac{2x}{1} = 2 \), which is wrong. Always check first that you really have \( \frac00 \) or \( \frac\infty\infty \), and check again before every repeated application.

Mistake 2 · Using the quotient rule The rule is \( \frac{f'}{\varphi'} \), the ratio of the two derivatives. It is not \( \left(\frac f\varphi\right)' = \frac{f'\varphi - f\varphi'}{\varphi^2} \).

Limitation · The limit of \(\frac{f}{\varphi}\) may exist when that of \(\frac{f'}{\varphi'}\) does not \[ \lim_{x\to\infty}\frac{x + \sin x}{x} = \lim_{x\to\infty}\Big(1 + \frac{\sin x}{x}\Big) = 1. \] But the rule gives \( \frac{1 + \cos x}{1} \), which oscillates between \(0\) and \(2\) and has no limit. The rule only says: if the right-hand limit exists, the left one equals it. If the right-hand limit does not exist, the rule says nothing, and you must try another method.

Limitation · The rule can go in circles For \( \lim_{x\to\infty}\frac{\sqrt{x^2 + 1}}{x} \), each application swaps numerator and denominator: \[ \frac{\sqrt{x^2+1}}{x} \to \frac{x/\sqrt{x^2+1}}{1} = \frac{x}{\sqrt{x^2+1}} \to \frac{\sqrt{x^2+1}}{x} \to \cdots \] Algebra is quicker: \( \frac{\sqrt{x^2+1}}{x} = \sqrt{1 + \frac1{x^2}} \to 1 \). L'Hôpital's rule is a tool, not a replacement for simplification.

6. The forms \(0\cdot\infty\) and \(\infty - \infty\)

6.1 The form \(0\cdot\infty\)

Rule If \( \lim_{x\to a}f_1(x) = 0 \) and \( \lim_{x\to a}f_2(x) = \infty \), write the product \( f_1(x)\cdot f_2(x) \) in the form \[ \frac{f_1(x)}{\dfrac{1}{f_2(x)}} \quad \left(\text{type } \tfrac00\right) \qquad \text{or} \qquad \frac{f_2(x)}{\dfrac{1}{f_1(x)}} \quad \left(\text{type } \tfrac\infty\infty\right), \] and find the limit as in Section 2.

Which form to choose? The one whose derivatives are simpler. A good rule of thumb: keep the logarithm (or inverse trigonometric function) in the numerator, because its derivative is simple, and move the power into the denominator.

Example 7 · \( \lim_{x\to 0^+} x\ln x \) Good choice (type \(\frac\infty\infty\)): \[ \lim_{x\to 0^+}\frac{\ln x}{1/x} = \lim_{x\to 0^+}\frac{1/x}{-1/x^2} = \lim_{x\to 0^+}(-x) = 0. \] Bad choice (type \(\frac00\)): \( \frac{x}{1/\ln x} \) gives \( \frac{1}{-1/(x\ln^2x)} = -x\ln^2x \), which is more complicated than what we started with. Both choices are legal; only one is useful. The result means: \(x \to 0\) wins against \(\ln x \to -\infty\).

6.2 The form \(\infty - \infty\)

Rule In the case \( \infty - \infty \), transform \( f_1(x) - f_2(x) \) into \[ f_1(x)\left[1 - \frac{f_2(x)}{f_1(x)}\right]. \] If \( \lim_{x\to a}\frac{f_2(x)}{f_1(x)} = 1 \), the indeterminacy has been reduced to the previous case, \( \infty\cdot 0 \).

(If that ratio tends to a number other than \(1\), there is nothing indeterminate left: the bracket tends to a nonzero number and the whole expression tends to \(\pm\infty\).) In practice, two other transformations are often quicker: bringing fractions to a common denominator, or multiplying by a conjugate expression.

Example 8 · Common denominator \[ \lim_{x\to 0}\left(\frac1x - \frac{1}{\sin x}\right) = \lim_{x\to 0}\frac{\sin x - x}{x\sin x} \;\left[\tfrac00\right] = \lim_{x\to 0}\frac{\cos x - 1}{\sin x + x\cos x} \;\left[\tfrac00\right] = \lim_{x\to 0}\frac{-\sin x}{2\cos x - x\sin x} = 0. \]

Example 9 · Factoring out, as in the book \( \lim_{x\to\infty}\big(\sqrt{x^2 + x} - x\big) \). Factor out \(x\) (here \( \frac{f_2}{f_1} = \frac{x}{\sqrt{x^2 + x}} \to 1 \)): \[ x\left[\sqrt{1 + \tfrac1x} - 1\right] = \frac{\sqrt{1 + t} - 1}{t} \quad (t = \tfrac1x \to 0^+) \quad \longrightarrow \quad \lim_{t\to 0}\frac{\frac{1}{2\sqrt{1+t}}}{1} = \frac12. \]

7. The forms \(1^\infty\), \(0^0\) and \(\infty^0\)

These arise for expressions of the form \( [f_1(x)]^{f_2(x)} \), where both the base and the exponent vary.

Rule · Take the logarithm Put \( y = [f_1(x)]^{f_2(x)} \). Then \[ \ln y = f_2(x)\ln f_1(x), \] and in all three cases this is of the form \(0\cdot\infty\):

\(1^\infty\): \( \infty\cdot\ln 1 = \infty\cdot 0 \);
\(0^0\): \( 0\cdot\ln 0 = 0\cdot(-\infty) \);
\(\infty^0\): \( 0\cdot\ln\infty = 0\cdot\infty \).

Find \( \lim\ln y = L \) as in Section 6; then \( \lim y = e^L \).

Don't forget the last step The most frequent error is to compute \(L = \lim\ln y\) and report \(L\) as the answer. The answer is \( e^L \).

Example 10 · The number \(e\): type \(1^\infty\) \( y = \big(1 + \frac1x\big)^x \), \( x \to \infty \). Then with \( t = \frac1x \): \[ \ln y = x\ln\Big(1 + \frac1x\Big) = \frac{\ln(1 + t)}{t} \;\left[\tfrac00\right] \longrightarrow \frac{1/(1 + t)}{1}\Big|_{t\to0} = 1. \] So \( \lim_{x\to\infty}\big(1 + \frac1x\big)^x = e^1 = e \). The intuitive guess "\(1\) to any power is \(1\)" is wrong: the base approaches \(1\) while the exponent explodes, and they balance at \(e\).

Example 11 · Type \(0^0\) \( \lim_{x\to 0^+}x^x \): \( \ln y = x\ln x \to 0 \) (Example 7), so \( \lim x^x = e^0 = 1 \).

Example 12 · Type \(\infty^0\) \( \lim_{x\to\infty}x^{1/x} \): \( \ln y = \frac{\ln x}{x} \to 0 \) (Example 4), so \( \lim x^{1/x} = 1 \).

Example 13 · A less obvious \(1^\infty\) \( \lim_{x\to 0}(\cos x)^{1/x^2} \): \[ \ln y = \frac{\ln\cos x}{x^2} \;\left[\tfrac00\right] = \frac{-\operatorname{tg}x}{2x} \longrightarrow -\frac12. \] So the limit is \( e^{-1/2} = \frac{1}{\sqrt e} \approx 0.607 \).

8. Comparing growth: logarithms, powers, exponentials

Examples 4 and 5 generalize to a hierarchy that is worth memorizing. For every \( a > 0 \) and \( b > 1 \), as \( x \to \infty \):

\[ \frac{\ln x}{x^a} \to 0, \qquad \frac{x^a}{b^x} \to 0. \]

In words:

\[ \ln x \;\ll\; x^a \;\ll\; b^x \;\ll\; x! \;\ll\; x^x. \]

Every power beats every logarithm (even \(\sqrt[100]{x}\) eventually beats \(\ln x\)), and every exponential beats every power. Once you know this table, many limits can be answered at a glance. The same hierarchy explains the convergence tests from the first posts of the series: why \( \sum\frac{n^{10}}{2^n} \) converges, and why \( \sum\frac{1}{n\ln n} \) diverges more slowly than the harmonic series.

9. Summary

Form	Transformation	Then
\( \frac00 \), \( \frac\infty\infty \)	none	\( \lim\frac{f}{\varphi} = \lim\frac{f'}{\varphi'} \)
\( 0\cdot\infty \)	\( f_1f_2 = \frac{f_1}{1/f_2} \) or \( \frac{f_2}{1/f_1} \)	\( \frac00 \) or \( \frac\infty\infty \)
\( \infty - \infty \)	\( f_1\big[1 - \frac{f_2}{f_1}\big] \), common denominator, or conjugate	\( \infty\cdot 0 \) or \( \frac00 \)
\( 1^\infty \), \( 0^0 \), \( \infty^0 \)	\( \ln y = f_2\ln f_1 \)	\( 0\cdot\infty \); answer \( e^{\lim\ln y} \)

Checklist before every application:

Is the form really \( \frac00 \) or \( \frac\infty\infty \)? If not, transform it first (or don't use the rule).
Differentiate numerator and denominator separately.
Simplify. Then check the form again before applying the rule a second time.
If \( \frac{f'}{\varphi'} \) has no limit, or the rule goes in circles, use another method (algebra, known limits, Taylor series).
For exponential forms, remember: the answer is \( e^{L} \), not \(L\).

10. Practice problems

Try each one before opening the solution. More solved problems from this topic are in Differential Calculus.

Problem 1: \( \displaystyle\lim_{x\to 0}\frac{e^x - 1 - x}{x^2} \)

\( \left[\tfrac00\right] \to \frac{e^x - 1}{2x} \left[\tfrac00\right] \to \frac{e^x}{2} \to \frac12 \).

Problem 2: \( \displaystyle\lim_{x\to 0}\frac{\operatorname{tg}x - x}{x - \sin x} \)

\( \frac{\frac{1}{\cos^2x} - 1}{1 - \cos x} = \frac{\sin^2x}{\cos^2x\,(1 - \cos x)} = \frac{(1 - \cos x)(1 + \cos x)}{\cos^2x\,(1 - \cos x)} = \frac{1 + \cos x}{\cos^2x} \to 2 \). (Simplifying after the first step avoids two more differentiations.)

Problem 3: \( \displaystyle\lim_{x\to\infty}\frac{\ln x}{\sqrt x} \)

\( \frac{1/x}{1/(2\sqrt x)} = \frac{2}{\sqrt x} \to 0 \).

Problem 4: \( \displaystyle\lim_{x\to 0^+}\sqrt x\,\ln x \)

Type \(0\cdot\infty\): \( \frac{\ln x}{x^{-1/2}} \to \frac{1/x}{-\frac12x^{-3/2}} = -2\sqrt x \to 0 \).

Problem 5: \( \displaystyle\lim_{x\to 1}\left(\frac{x}{x - 1} - \frac{1}{\ln x}\right) \)

Common denominator: \( \frac{x\ln x - x + 1}{(x - 1)\ln x} \left[\tfrac00\right] \to \frac{\ln x}{\ln x + 1 - \frac1x} \left[\tfrac00\right] \to \frac{1/x}{\frac1x + \frac1{x^2}} \to \frac{1}{2} \).

Problem 6: \( \displaystyle\lim_{x\to\infty}\Big(1 + \frac ax\Big)^x \)

\( \ln y = x\ln\big(1 + \frac ax\big) = \frac{\ln(1 + at)}{t} \) with \(t = \frac1x\), which tends to \(a\). So the limit is \( e^a \).

Problem 7: \( \displaystyle\lim_{x\to 0^+}x^{\sin x} \)

Type \(0^0\): \( \ln y = \sin x\ln x = \frac{\sin x}{x}\cdot x\ln x \to 1\cdot 0 = 0 \). The limit is \( e^0 = 1 \).

Problem 8: Why can't L'Hôpital's rule be used for \( \displaystyle\lim_{x\to\infty}\frac{x - \cos x}{x + \cos x} \)? Find the limit anyway.

The rule gives \( \frac{1 + \sin x}{1 - \sin x} \), which has no limit (and is even undefined where \(\sin x = 1\)). Divide by \(x\) instead: \( \frac{1 - \frac{\cos x}{x}}{1 + \frac{\cos x}{x}} \to 1 \).

Final thoughts

L'Hôpital's rule turns a question about values ("what is \(\frac00\)?") into a question about rates ("which side is changing faster?"). It comes straight from Cauchy's mean value theorem, works for \( \frac00 \) and \( \frac\infty\infty \), and with a little algebra handles the other five indeterminate forms as well. Used carelessly, it gives wrong answers; used with the checklist above, it is the fastest route to most limits you will meet in calculus.

Related posts: the mean value theorems (the proof of the rule), the differential (why functions look linear near a point), and power series (an alternative way to compute the same limits).