# Chapter 14: L'Hôpital's Rule and Overview of Limits

> Master the indeterminate forms 0/0 and ∞/∞, convert ∞−∞, 0·∞ and exponential forms with algebra and logarithms, and see the whole limit toolkit in one map.

**Canonical:** https://calculus.kitzo.work/chapter/lhopital  
**Level:** Advanced  
**Estimated time:** 40 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- 0/0 — Type A
- ∞/∞ — Type A
- ∞ − ∞ — Type B
- 0 · ∞ — Type B
- 1^∞, 0^0, ∞^0 — Type C
- Summary of all forms
- The master plan for limits

**Why this matters**

Limits are the gatekeeper to continuity, derivatives, integrals, and every theorem after them. Most limit problems collapse into one of a handful of indeterminate forms. L'Hôpital's rule disposes of the fraction forms in one stroke, and logarithm tricks convert the exponential ones. Learn the conversion playbook once and entire families of scary-looking limits become routine.

## 1 · The rule and why it works

**L'Hôpital's rule.** If f(x) and g(x) both tend to 0 as x → a (or both → ±∞), and f′/g′ has a limit as x → a, then

lim f(x)/g(x) = lim f′(x)/g′(x)

Why it works: near a, both functions are tiny (or gigantic), so their ratio is governed by their local slopes. Formally, Cauchy's mean value theorem produces ξ between x and a with f(x)/g(x) = f′(ξ)/g′(ξ), and as x → a the point ξ is squeezed to a as well. The derivative ratio then inherits the same limit. Requirements: the form is 0/0 or ∞/∞, and the new limit exists.

## 2 · Type A — 0/0

**Worked example 1**

### Three 0/0 limits

- **Polynomials:** lim x→1 (x² − 1)/(x − 1). Direct substitution is 0/0. Differentiate: (2x)/(1) → 2. (Or just factor: (x+1)(x−1)/(x−1) = x+1 → 2 — factoring is always a fine first move.)

- **Trig:** lim x→0 sin(3x)/x. Substitution gives 0/0. Differentiate: cos(3x)·3/1 → 3. (Or use sin u/u → 1 with u = 3x.)

- **Exp/log:** lim x→0 (eˣ − 1)/x. Substitution gives 0/0. Differentiate: eˣ/1 → 1. This is the definition-in-disguise of eˣ being its own derivative.

Each application requires re-checking that the new fraction is still 0/0 or ∞/∞ before differentiating again — repeated applications are legal, e.g. lim x→0 (1 − cos x)/x²: first pass gives sin x / 2x (still 0/0), second gives cos x / 2 → 1/2.

## 3 · Type A — ∞/∞

**Worked example 2**

### Growth dominance: two ∞/∞ limits

- **Log vs polynomial:** lim x→∞ ln x / x. Both → ∞. Differentiate: (1/x)/1 = 1/x → 0. Logarithms lose to powers, always.

- **Polynomial vs exponential:** lim x→∞ x²/eˣ. Both → ∞. Differentiate twice: 2x/eˣ (still ∞/∞), then 2/eˣ → 0. Exponentials beat every polynomial, and l'Hôpital proves it mechanically.

## 4 · Type B1 (∞ − ∞) and Type B2 (0 · ∞)

Neither of these can be differentiated directly — l'Hôpital only handles fractions. Convert first:

- **∞ − ∞:** combine into one fraction. Example: lim x→0⁺ (1/x − 1/sin x) = lim (sin x − x)/(x sin x). This is 0/0; differentiate: (cos x − 1)/(sin x + x cos x), still 0/0; differentiate again → −sin x/(2cos x − x sin x) → 0.

- **0 · ∞:** flip one factor. lim x→0⁺ x·ln x → form 0 · (−∞). Rewrite as ln x/(1/x) → −∞/∞. Differentiate: (1/x)/(−1/x²) = −x → 0. So x·ln x → 0.

## 5 · Type C — 1^∞, 0^0, ∞^0

**Worked example 3**

### Exponential-form limits via logarithms

- **Setup:** let y = f(x)^g(x). Take logs: ln y = g(x)·ln f(x) — a Type B2 form. Find L = lim g ln f, then the answer is e^L.

- **Family classic:** lim x→∞ (1 + 1/x)^x. ln y = x·ln(1 + 1/x) → form 0 · ∞. Rewrite: ln(1 + 1/x)/(1/x) → 0/0. Differentiate: [1/(1 + 1/x)·(−1/x²)] / (−1/x²) = 1/(1 + 1/x) → 1. So ln y → 1 and y → e.

- **Zero to the zero:** lim x→0⁺ x^x. ln y = x·ln x → 0 (computed in Type B2). So y → e⁰ = 1.

## 6 · Summary of forms and the master plan

- **Type A (direct):** 0/0 and ∞/∞ → differentiate numerator and denominator; repeat if needed.

- **Type B1:** ∞ − ∞ → algebra into a single fraction.

- **Type B2:** 0 · ∞ → reciprocate one factor to make a fraction.

- **Type C:** 1^∞, 0^0, ∞^0 → take ln, solve, exponentiate back.

- **Not indeterminate:** k/0 (→ ±∞ or DNE), 0/∞ = 0, ∞ + ∞ = ∞, 1^0 = 1. Plug in first — many limits scream their answer.

**When l'Hôpital does NOT apply:** when the form is not indeterminate (differentiating a perfectly good quotient just produces a wrong derivative and a wrong answer); when the derivative ratio's limit does not exist (try another method); or when f′/g′ oscillates endlessly. Also remember l'Hôpital is heavy machinery — if factoring, conjugates, or squeeze theorem settle it in two lines, use them.

**The master plan — run this flowchart in order:**

- **Plug in.** If you get a finite number, ±∞, or a one-sided mismatch — done, or report DNE.

- **Simplify.** Factor and cancel, rationalize with a conjugate, or split the expression using algebra.

- **Known forms.** Use standard limits: sin x/x → 1, (eˣ − 1)/x → 1, (1 − cos x)/x → 0, ln(1 + x)/x → 1.

- **Indeterminate?** Convert to 0/0 or ∞/∞ (Types B and C via §4–5), then l'Hôpital. Repeat until resolved.

- **Squeeze theorem** when the expression is trapped between two functions with the same limit (sin x/x family, bounded · → 0).

- **Dominance hierarchy** at ±∞: logs < powers < exponentials < factorials.

- **Still stuck?** Numerical probe near the point, or series expansion for the stubborn transcendental mix.

**Watch out**

L'Hôpital is not 'differentiate top and bottom of any fraction.' Apply it to (x + 1)/(x + 2) as x → 0 and you would report 1/1 = 1 — but plain substitution gives 1/2, the true answer, because the form is not indeterminate. Verify the form first, every single time, before touching a derivative.

**Chapter formula sheet**

L'Hôpital: lim f/g = lim f′/g′ (for 0/0 or ∞/∞, when the latter limit exists)
∞ − ∞ → single fraction; 0 · ∞ → reciprocate one factor
Type C: y = f^g, ln y = g·ln f, then answer = e^(lim ln y)
Standard limits: sin x/x → 1, (eˣ − 1)/x → 1, ln(1 + x)/x → 1
Dominance: ln x ≪ x^k ≪ eˣ (x → ∞)
Squeeze: g ≤ f ≤ h, g and h share a limit ⇒ f shares it

## Common mistakes

- Applying l'Hôpital to a fraction that is not 0/0 or ∞/∞ — the single most common error on exams.

- Differentiating top and bottom as a product or quotient — l'Hôpital means separate derivatives, never the quotient rule.

- Declaring a limit DNE because f′/g′ oscillates, without trying factoring or the squeeze theorem.

- Forgetting to exponentiate back after a Type C logarithm detour — the answer is e^L, not L.

- Splitting an ∞ − ∞ difference into limits that are each ±∞ and writing '∞ − ∞ is undefined.'

- Misreading 1^∞ as certainly equal to 1 — it is genuinely indeterminate and demands the ln treatment.

## Practice exercises

1. Which of these expressions is indeterminate, so that l'Hôpital's rule might be needed?

- A. 3 / 0⁺
- B. 0 / 0
- C. ∞ / 2
- D. 1 + 0

2. Evaluate lim x→0 sin(3x) / x.

3. Evaluate lim x→0 (e^x − 1) / x.

4. Evaluate lim x→0 (1 − cos x) / x². Answer as a fraction or decimal, e.g. 1/2 or 0.5.

5. The limit lim x→0⁺ x·ln x has the form 0 · (−∞). Which rewrite makes it l'Hôpital-ready?

- A. ln x / (1/x) — now −∞/∞
- B. x / ln x — now 0/−∞
- C. ln x · x — no change needed
- D. x^2 · ln x / x

6. Evaluate lim x→∞ x² / e^x.

7. Evaluate lim x→∞ (1 + 2/x)^x. Answer exactly, e.g. e^2.

8. For which of these limits does l'Hôpital's rule NOT apply?

- A. lim x→0 sin x / x (form 0/0)
- B. lim x→0 (x + 1) / (x + 2)
- C. lim x→∞ ln x / x (form ∞/∞)
- D. lim x→0 (e^x − 1) / x (form 0/0)

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