This chapter closes on a tool that isn't integration at all, but earns its place here anyway. A limit that direct substitution turns into \frac{0}{0} or \frac{\infty}{\infty} doesn't announce its value. Those forms are called INDETERMINATE because the same form can resolve to any number, or to infinity, or fail to exist, depending on which functions produced it. L'Hopital's rule replaces the guesswork with a single, mechanical move.
\frac{0}{0} isn't one number pretending to be many. It's genuinely ambiguous, since the actual value depends entirely on how fast the numerator and denominator are each racing toward zero. L'Hopital's rule settles that race by comparing their SPEEDS, their derivatives, instead of their positions, the original functions. That's exactly the extra information direct substitution threw away. Comparing speeds instead of positions is exactly what turns an unresolvable ratio into one with a definite answer. This only works when both racers are actually headed to 0 or both to infinity together. Plugging the rule into a limit that already has a plain, finite value produces nonsense, since there was never a race to settle in the first place.
L'Hopital's rule is a mechanical way to resolve an indeterminate form. An indeterminate form is a limit expression, like \frac{0}{0}, whose value is not decided by direct substitution alone. The same form can resolve to any number, infinity, or fail to exist, depending on which functions produced it. The rule replaces that guesswork with a single move: differentiate the numerator and denominator separately, then take the limit of that new ratio. The rule only applies to the two indeterminate forms it names: \frac{0}{0} and \frac{\pm\infty}{\pm\infty}. Applying it to anything else, a form that already evaluates, or a different indeterminate shape like 0\cdot\infty or \infty-\infty, gives a meaningless answer, not a wrong one you could sanity-check away. Those other forms need to be algebraically rewritten into a \frac00 or \frac{\infty}{\infty} shape FIRST before the rule is even eligible to apply. That classification step is what this section trains directly. Misclassifying the form is the far more common error, more common than misapplying the derivative once the form is confirmed.
Theorem
L'Hopital's Rule
Recalled from Calculus I — Differential & Integral Calculus · Indeterminate forms and L'Hopital's Rule — first taught there.
Suppose f and g are differentiable near a (except possibly at a), g'(x) \neq 0 near a, and \lim_{x\to a}\frac{f(x)}{g(x)} \text{ is of the indeterminate form } \frac{0}{0} \text{ or } \frac{\pm\infty}{\pm\infty}. Then \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}, provided the limit on the right exists (or is \pm\infty). The rule also holds for x \to a^+, x \to a^-, or x \to \pm\infty.
The \frac{0}{0} case follows from the Generalized (Cauchy) Mean Value Theorem: for x near a, there is a c between a and x with \dfrac{f(x)-f(a)}{g(x)-g(a)} = \dfrac{f'(c)}{g'(c)}. Since f(a)=g(a)=0 in the \frac{0}{0} case, the left side is exactly \frac{f(x)}{g(x)}; letting x \to a forces c \to a too, giving the limit of f'/g'. The \frac{\infty}{\infty} case uses the same idea with a more careful algebraic argument and is not reproduced here.
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Classify before you differentiate
Mark whether \dfrac{x+2}{x-1} is positive or negative on each interval -- the same interval-by-interval sign reasoning shows up constantly once limits and integrals get more complicated, so it's worth having automatic.
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Worked example
Applying L'Hopital's rule once
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1
Evaluate \displaystyle\lim_{x\to 0}\frac{\sin x}{x} using L'Hopital's rule (rather than the Squeeze-Theorem argument from Calculus I).
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2
Direct substitution gives \frac{\sin 0}{0} = \frac{0}{0}, an eligible indeterminate form.
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3
Differentiate numerator and denominator SEPARATELY (not as a quotient rule): \frac{d}{dx}[\sin x] = \cos x and \frac{d}{dx}[x] = 1.
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4
\lim_{x\to0}\dfrac{\cos x}{1} = \cos 0 = 1. \boxed{\lim_{x\to0}\dfrac{\sin x}{x} = 1}, matching the earlier Squeeze-Theorem result by a completely different route.
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Where this goes next
Improper integrals and series, later in this course
Indeterminate forms resurface constantly: evaluating an improper integral's limit, and comparing growth rates deep in the series chapters. L'Hopital's rule is the tool that settles most of them cleanly.
Check your understanding (counts toward your progress)
Evaluate \displaystyle\lim_{x\to0}\dfrac{e^x-1}{x} using L'Hopital's rule.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\frac00 form; differentiate top and bottom — \frac{e^x}{1}\to e^0 = 1.
Correct answer
Also accepted:
Why
Confirm this is a \frac00 form, then differentiate the numerator and denominator separately.
To apply L'Hopital's rule to \displaystyle\lim_{x\to0^+} x\ln x (form 0\cdot(-\infty)), what must happen first?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
0\cdot(-\infty) isn't one of the two eligible forms — applying the rule directly here is a misuse of it.
It CAN be evaluated — just not with L'Hopital's rule until it's rewritten into an eligible form first.
Multiplying by x changes the expression rather than reshaping it into a valid quotient form.
Why
Rewriting x\ln x as \dfrac{\ln x}{1/x} turns 0\cdot(-\infty) into \dfrac{-\infty}{\infty}, an eligible form.
Rewrite the product as a fraction so the form becomes \frac{-\infty}{\infty} instead of 0\cdot(-\infty).
What indeterminate form does \displaystyle\lim_{x\to 0^+} 2x^{1}\ln x have, before any simplification?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Substituting the limit point directly into \displaystyle\lim_{x\to 0^+} 2x^{1}\ln x produces the form 0\cdot\infty — recognizing which form is the first step before choosing a technique (L'Hopital's rule, algebraic rewriting, etc.).
What indeterminate form does \displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+9x}-x\right) have, before any simplification?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Substituting the limit point directly into \displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+9x}-x\right) produces the form \infty-\infty — recognizing which form is the first step before choosing a technique (L'Hopital's rule, algebraic rewriting, etc.).