# Chapter 3: Introduction to Limits

> What a limit really is, one-sided limits, the ways limits fail to exist, limits at ±∞, vertical and horizontal asymptotes done correctly, and the Sandwich Principle with sin x / x near 0.

**Canonical:** https://calculus.kitzo.work/chapter/limits-intro  
**Level:** Starter  
**Estimated time:** 35 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- The limit idea
- Left & right-hand limits
- When limits do not exist
- Limits at ±∞
- Asymptotes, correctly
- The Sandwich Principle

**Why this matters**

Limits are the one genuinely new idea in calculus. Derivatives are limits of slopes; integrals are limits of sums. Master this chapter and the rest of the course is technique — skip it and everything else is ritual without meaning.

## 1 · The limit idea and its notation

The expression `lim x→a f(x) = L` says: *as x gets arbitrarily close to a, f(x) gets arbitrarily close to L*. Two things are pointedly irrelevant: what f(a) equals (it may not even exist), and what happens exactly at a. Limits describe the journey, not the destination.

The rigorous version uses ε and δ: for every ε > 0 you can name a δ > 0 so that 0 < |x − a| < δ forces |f(x) − L| < ε. You rarely need to produce δ by hand, but the logic — *every tolerance, however small, can be met* — is the soul of the definition.

**Worked example**

**Example 1 — Find lim x→2 (x² − 4)/(x − 2).**

Step 1: plugging in gives 0/0 — inconclusive, not an answer.

Step 2: factor: (x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2.

Step 3: the limit only cares about x near 2, not at 2, so lim x→2 (x + 2) = 4.

The graph has a removable hole at (2, 4) — the limit is 4 even though f(2) is undefined.

## 2 · Left- and right-hand limits

`lim x→a⁺ f(x)` approaches from the right (x > a); `lim x→a⁻ f(x)` from the left (x < a). The two-sided limit exists **if and only if** both one-sided limits exist and agree. Piecewise functions are judged at their seams exactly this way.

**Worked example**

**Example 2 — f(x) = x + 1 for x < 1, and f(x) = 4 − x for x ≥ 1. Find lim x→1 f(x).**

Step 1: from the left: lim x→1⁻ (x + 1) = 2.

Step 2: from the right: lim x→1⁺ (4 − x) = 3.

Step 3: 2 ≠ 3, so the two-sided limit does not exist. The graph jumps from height 2 to height 3 at x = 1.

## 3 · When limits fail to exist

Three classic failures:

  - **Jump discontinuity** — the one-sided limits exist but disagree (Example 2).

  - **Oscillation** — f(x) = sin(1/x) near 0 swings between −1 and 1 infinitely often in any window around 0; it never settles, so no L exists.

  - **Blow-up** — f(x) = 1/x² near 0 rockets past every number; the function grows without bound (we write lim x→0 1/x² = ∞ as shorthand, and ∞ is not a real number, so the limit fails to exist in the strict sense).

**Watch out**

Writing lim x→0 1/x = ∞ is wrong notation. The right-hand limit is ∞ but the left-hand limit is −∞; they disagree, and ±∞ are not numbers. Only write ∞ when one side is unbounded and you know which direction, or 1/x²-style collapse where both sides agree on the infinity.

## 4 · Limits at ±∞ and asymptotes, correctly

`lim x→∞ f(x) = L` means f(x) settles near L as x grows without bound. For rational functions, compare degrees: same degree → ratio of leading coefficients; denominator higher degree → 0; numerator higher degree → unbounded.

**Vertical asymptote** at x = a: a one-sided (or two-sided) limit there is ±∞ — the function *blows up* near a. **Horizontal asymptote** y = L: lim x→∞ f(x) = L or lim x→−∞ f(x) = L — the function *levels off* far out. An asymptote is a statement about a limit, never about a single point, and graphs may cross horizontal asymptotes without invalidating them.

## 5 · The Sandwich Principle (squeeze theorem)

If g(x) ≤ f(x) ≤ h(x) near a (except possibly at a) and lim g = lim h = L, then lim f = L — the function is squeezed into L. This is how the course's most famous limit is proved.

**Worked example**

**Example 3 — Show lim x→0 sin x / x = 1.**

Step 1: on the unit circle, for small x > 0, comparing areas gives sin x < x < tan x.

Step 2: divide by sin x: 1 < x/sin x < 1/cos x, so cos x < sin x / x < 1.

Step 3: both cos x and 1 squeeze to 1 as x → 0. By the Sandwich Principle, lim x→0 sin x / x = 1.

The same bound works for x < 0 by symmetry, since sin x / x is even. This limit is the engine behind every derivative of sin x, so it is worth seeing once.

More generally, if |f(x)| ≤ M·|x|^p near 0 with p > 0 you know instantly the limit is 0; and a function trapped between two identical limits has nowhere else to go.

**Chapter formula sheet**

lim x→a f(x) = L ⇔ lim x→a⁻ f(x) = lim x→a⁺ f(x) = L

lim x→0 sin x / x = 1 · lim x→0 (1 − cos x) / x = 0

Rational at ∞: same degree → ratio of leading coefficients · denominator higher → 0

Sandwich: g ≤ f ≤ h, lim g = lim h = L ⇒ lim f = L

ε–δ: ∀ ε > 0 ∃ δ > 0 with 0 < |x − a| < δ ⇒ |f(x) − L| < ε

## Common mistakes

  - Plugging in, getting 0/0, and writing "the limit is 0/0" — 0/0 means *keep working*.

  - Assuming the two-sided limit exists because one side exists — both sides must exist and agree.

  - Calling ∞ a value of the limit; it signals unbounded behavior, not a real-number answer.

  - Confusing horizontal asymptotes with hard ceilings — graphs can cross them.

  - Declaring lim x→0 sin(1/x) = 0 because sin equals 0 infinitely often; it never settles.

  - Using L'Hôpital's rule before its preconditions (differentiable pieces, genuine 0/0 or ∞/∞) are met.

## Practice exercises

1. lim x→2 (x² − 4)/(x − 2) equals:

- A. 0
- B. 2
- C. 4
- D. does not exist

2. Compute lim x→3 (2x + 1).

3. Evaluate lim x→0 sin(3x) / x.

4. For f(x) = (x − 1)/|x − 1|, the limit lim x→1 f(x):

- A. equals 1
- B. equals −1
- C. equals 0
- D. does not exist

5. Compute lim x→∞ (3x² + x) / (x² − 7).

6. lim x→0 sin(1/x):

- A. equals 0
- B. equals 1
- C. does not exist (oscillation)
- D. equals ∞

7. Evaluate lim x→∞ 5/(x + 2). Enter a number.

8. Which statement about the Sandwich Principle is correct?

- A. If g ≤ f ≤ h and g, h both tend to L, then f tends to L
- B. If f is between g and h, f must equal the average of g and h
- C. It only applies to polynomials
- D. It requires f to be differentiable

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