Lesson 12: L'Hôpital's Rule and Indeterminate Forms

A cluster of limit forms resist the methods of Lessons 1–2: 0/0 and ∞/∞ again, but also 0·∞, ∞ − ∞, 1^∞, 0^0, ∞^0. L'Hôpital's rule is the heavy tool for the first two forms: if f and g both tend to 0 (or both to ±∞), then lim f/g = lim f'/g', provided the latter exists. The other forms are not new techniques at all — they are algebra problems whose job is to convert themselves into 0/0 or ∞/∞ so the rule can bite. This lesson is about those conversions, and about the discipline of checking the form before differentiating anything.

The rule at work

Worked Example 1. lim(x→0) sin(3x)/x. Form 0/0. Differentiate top and bottom separately: lim 3cos(3x)/1 = 3. (The pattern lim sin(ax)/x = a is worth owning outright.) Worked Example 2 (apply twice). lim(x→0) (e^x − 1 − x)/x^2. Form 0/0; one application gives (e^x − 1)/(2x), still 0/0; apply again: e^x/2 → 1/2. Nothing about the rule limits you to one pass — but each pass requires rechecking that the form is still indeterminate. Worked Example 3 ( ∞/∞ and growth rates). lim(x→∞) x/e^x: one application gives 1/e^x → 0. Compare the family: lim (ln x)/x = lim (1/x)/1 = 0, and lim x^2/e^x = 0 after two passes. The hierarchy to internalise: logarithms grow slower than any power, which grows slower than any exponential. Most growth-rate questions are decided by this sentence alone.

Converting the other forms

• 0 · ∞ → quotient. lim(x→0+) x ln x: rewrite as ln x/(1/x) (form −∞/∞), apply the rule: (1/x)/(−1/x^2) = −x → 0. Exponentials and logs beat the small factor to the punch — a result with real content: x shrinks faster than ln x explodes.
• ∞ − ∞ → common denominator. Worked Example 4. lim(x→1) ( 1/(x − 1) − 2/(x^2 − 1) ). Combine over (x − 1)(x + 1): numerator (x + 1) − 2 = x − 1, so the expression is (x − 1)/((x − 1)(x + 1)) = 1/(x + 1) → 1/2. No L'Hôpital needed once the algebra is done — conversion first, calculus second.
• Powers 1^∞, 0^0, ∞^0 → take logarithms. Worked Example 5. lim(x→0+) (1 + x)^(1/x). Let L be the limit and take ln: ln L = lim (ln(1 + x))/x, form 0/0; L'Hôpital gives lim (1/(1 + x))/1 = 1, so ln L = 1 and L = e. You have just met the limit definition of e hiding inside an indeterminate power.

A logical caveat, honestly stated

Using L'Hôpital on lim(x→0) sin x/x is circular if the derivative of sin was itself established using that very limit (it was, in most courses). The resolution: in a rigorous development, sin x/x = 1 is proved geometrically first (the squeeze theorem), and then L'Hôpital is legitimate everywhere else. Know the caveat; it distinguishes understanding the tool from merely swinging it. A second caution: the rule can loop — for lim(x→∞) x/sqrt(x^2 + 1), differentiating returns an expression whose limit leads back to the original. When the rule cycles, switch methods (here: divide top and bottom by x to get 1).

Application: comparing growth in the wild

A startup's user base grows like U(t) = t^3 (thousands) while a rival's grows like R(t) = 2^t. Which dominates eventually? lim t^3/2^t = 0 (three L'Hôpital passes, or the hierarchy sentence), so the exponential rival wins in the long run despite trailing early — solve t^3 = 2^t numerically and the crossover sits near t ≈ 9.9 (check t = 10: 1000 vs 1024). Investment, epidemiology, and algorithm analysis (n^2 vs 2^n operations!) all reduce to this comparison. The concepts-first moral: eventual behaviour is a limit, and limits are decided by growth hierarchy, not by who is ahead now.

Common mistakes

• Applying the rule when the form is not indeterminate. Wrong: lim(x→0) (x + 1)/(x + 2) "by L'Hôpital" = 1/1 = 1. Right: direct substitution gives 1/2; the rule's hypotheses fail (numerator tends to 1, not 0), so its conclusion is worthless here. Check the form first, every time.
• Differentiating the quotient instead of top and bottom separately. L'Hôpital replaces f/g by f'/g' — it is not the quotient rule. Writing (f'g − fg')/g^2 is the signature error.
• Giving up after one pass. (e^x − 1 − x)/x^2 needs two passes; after each, recheck the form. Stopping early and declaring "still 0/0, so the limit does not exist" confuses the method's progress with the answer.
• Forgetting to exponentiate back. In 1^∞ problems the logarithm's limit is not the answer: ln L = 1 means L = e. Half of all errors in this family are this final step skipped.

Exam Experience

1. lim(x→0) (e^x − 1)/x equals

• (A) 0
• (B) 1
• (C) e
• (D) The limit does not exist.

Answer: (B). Form 0/0; L'Hôpital gives e^x/1 → 1. (This is also the derivative of e^x at 0 in disguise — the definition from Lesson 3.)

2. lim(x→∞) (3x^2 + 1)/(e^x) equals

• (A) 3
• (B) 1
• (C) 0
• (D) ∞

Answer: (C). Two L'Hôpital passes (or the growth hierarchy): (6x)/(e^x), then 6/e^x → 0. Exponentials outgrow every polynomial.

3. lim(x→0+) (1 + 2x)^(1/x) equals

• (A) e
• (B) e^2
• (C) 2
• (D) 1

Answer: (B). ln of the expression is ln(1 + 2x)/x → (by L'Hôpital) 2/(1 + 2x) → 2, so the limit is e^2. Worked Example 5 is the same machine with coefficient 1.

Free-response question. (a) Evaluate lim(x→0) (sin x − x)/x^3. (b) Evaluate lim(x→0+) x^2 ln x. (c) A student evaluates lim(x→2) (x^2 − 3)/(x + 1) by L'Hôpital and obtains 2x/1 → 4. Identify the error and give the correct value.

Scoring guideline.

• (a) 2 points. Form 0/0. 1 point for two correct passes: (cos x − 1)/(3x^2), still 0/0, then (−sin x)/(6x), still 0/0 — a third pass (or recognising sin x/x → 1) gives −cos x/6 → −1/6. Award the second point for the value −1/6 with the multi-pass working visible; an answer after only two passes ("still indeterminate") earns 1.
• (b) 1 point. Convert: ln x/(1/x^2) → (1/x)/(−2/x^3) = −x^2/2 → 0.
• (c) 1 point. The form at x→2 is (4 − 3)/(3) = 1/3, not indeterminate, so L'Hôpital does not apply; the correct value by substitution is 1/3. (The student's 4 is the classic hypothesis-skipping error.)

Going deeper: three more forms, and a classification drill

A disguised derivative quotient. lim(x→0) (e^(2x) − e^(−2x))/(2x): form 0/0. L'Hôpital: numerator derivative 2e^(2x) + 2e^(−2x) → 4, denominator derivative 2, limit 2. (It is also (f(x) − f(−x))/(2x) for f = e^(2x) in disguise — Lesson 3's definition, symmetric version.) A power that is not what it seems. lim(x→∞) x^(1/x): the base explodes, the exponent dies — form ∞^0. Take ln: (ln x)/x → 0 by the hierarchy, so the limit is e^0 = 1. Roots grow so slowly that even x itself, rooted x-deep, is crushed to 1. Three honest passes. lim(x→0) (tan x − x)/x^3: 0/0 three times in succession — (sec^2 x − 1)/(3x^2), then (2 sec^2 x tan x)/(6x), then one more pass — yields 1/3. (Lesson 16 explains the shortcut: tan x = x + x^3/3 + ..., so the numerator is essentially x^3/3. Series and L'Hôpital are two views of the same local behaviour.) Classification drill — name the form before touching anything: lim(x→1) (x^2 + 1)/(x − 1): not indeterminate at all — the numerator tends to 2, so the one-sided limits are ±∞ by direct sign analysis; L'Hôpital is simply inapplicable. lim(x→0) (2^x − 1)/x: 0/0, and the rule gives (2^x ln 2)/1 → ln 2. lim(x→∞) (x + sin x)/x: ∞/∞, but try the rule and it returns (1 + cos x)/1, which has no limit — the rule's conclusion fails because lim f'/g' does not exist, while the original limit plainly equals 1 (divide through by x: 1 + (sin x)/x → 1). The hypotheses of L'Hôpital include "lim f'/g' exists" — this example is why.

Going further: differences of roots, and 0^0 forms

Indeterminate differences. lim(x→∞) ( sqrt(x^2 + x) − sqrt(x^2 − x) ): form ∞ − ∞. The conjugate converts it to a quotient: multiply by ( sqrt(x^2 + x) + sqrt(x^2 − x) ) over itself; the numerator becomes (x^2 + x) − (x^2 − x) = 2x, so the limit is lim 2x/( sqrt(x^2 + x) + sqrt(x^2 − x) ) = 2x/(x + x) = 1 after dividing through by x. Rationalisation — Lesson 1's trick — is the standard first move on root differences at every scale. The form 0^0. lim(x→0+) (sin x)^x: the base dies, the exponent dies. Logarithm: x·ln(sin x) = ln(sin x)/(1/x), an ∞/∞ for L'Hôpital: derivative ratio (cos x/sin x)/(−1/x^2) = −x^2 cos(x)/sin(x) → 0 (because x/sin x → 1). So the log of the limit is 0 and the limit is e^0 = 1. Two of the course's tools (the ln conversion and the sin(x)/x benchmark from Lesson 1) compose into one answer — indeterminate forms are where the toolbox stops being a list and becomes a kit.

One more two-pass exercise to cement the rhythm: lim(x→0) (1 − cos x)/(x sin x). Form 0/0; differentiate: sin x/(sin x + x cos x) — still 0/0 at 0 (the denominator vanished too). Differentiate again: cos x/(2 cos x − x sin x) → 1/(2 − 0) = 1/2. Two lessons in one line: a single application of the rule is not a commitment — you may apply it as often as the form persists — and each pass should be *simplified* before the next, or the algebra compounds beyond control. (The series route of Lesson 16 agrees: numerator ≈ x^2/2, denominator ≈ x^2.)