# Chapter 4: Solving Limit Problems Involving Polynomials

> The workhorse toolkit for hands-on limit problems: factor and cancel at finite points, the conjugate trick for square roots, divide by the highest power at infinity, degree comparison, and absolute-value and one-sided infinite limits.

**Canonical:** https://calculus.kitzo.work/chapter/limits-polynomials  
**Level:** Core  
**Estimated time:** 50 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Factor and cancel
- Conjugate trick
- Divide by highest power
- Degree comparison at ±∞
- Absolute-value limits
- One-sided infinite limits

**Why this matters**

Ninety percent of the limits you will actually meet in calculus, precalculus review and exams are algebraic: ratios of polynomials, or polynomials dressed in radicals and absolute values. This chapter is the toolkit that turns "0/0, do not know what to do" into a two-line computation.

## 1 · As x → a: factor and cancel

When direct substitution gives 0/0 in a rational function, the zero is a shared factor wearing a disguise. Factor numerator and denominator, cancel the common factor, then substitute.

**Worked example**

**Example 1 — Find lim x→3 (x² − 9)/(x² − x − 6).**

Step 1: substitute: 0/0 — factor instead.

Step 2: numerator (x − 3)(x + 3); denominator (x − 3)(x + 2).

Step 3: cancel the (x − 3) — legal because the limit never touches x = 3 itself.

Step 4: lim x→3 (x + 3)/(x + 2) = 6/5.

If nothing cancels after full factoring, you are looking at a genuine infinite limit or a case where one side goes to ∞ and the other to −∞ (Section 5).

## 2 · Square roots: multiply by the conjugate

Radicals produce 0/0 through a *difference of roots*. The cure is the conjugate: multiply top and bottom by the mirror expression with the sign flipped, then use (A − B)(A + B) = A² − B².

**Worked example**

**Example 2 — Find lim x→0 (√(x + 4) − 2)/x.**

Step 1: direct substitution: (2 − 2)/0 = 0/0.

Step 2: multiply by (√(x + 4) + 2)/(√(x + 4) + 2):

((x + 4) − 4) / (x·(√(x + 4) + 2)) = x / (x·(√(x + 4) + 2)).

Step 3: cancel x: 1/(√(x + 4) + 2).

Step 4: substitute x = 0: 1/(2 + 2) = 1/4.

## 3 · As x → ±∞: divide by the highest power

At infinity, no single term wins by magnitude except the highest-power one. Divide every term top and bottom by xⁿ (n = top degree, adjusting if needed) and watch the lower terms die as 1/x → 0.

**Worked example**

**Example 3 — Find lim x→∞ (2x² + 3x)/(5x² − 7).**

Step 1: highest power is x² — divide: (2 + 3/x) / (5 − 7/x²).

Step 2: let x → ∞: 3/x → 0 and 7/x² → 0.

Step 3: lim = 2/5.

For radicals at infinity, pull the root out: √(9x² + 1) = |x|√(9 + 1/x²). Note the absolute value — for x → −∞, √(x²) = −x, not x.

## 4 · The degree comparison rule

For rational f(x) = p(x)/q(x) as x → ±∞, compare degrees:

  - deg p < deg q → limit is **0** (horizontal asymptote y = 0).

  - deg p = deg q → limit is the **ratio of leading coefficients** (horizontal asymptote y = a/b).

  - deg p > deg q → the limit is **±∞**; by exactly 1 degree the asymptote is a slant line, otherwise it is parabolic/cubic growth.

This rule is a shortcut, not a substitute for understanding — but at speed, it reads the answer off the top line of the problem.

## 5 · Absolute values and one-sided infinite limits

Absolute values split the domain at their zero. |x − a| equals (x − a) when x ≥ a and −(x − a) when x < a. So a limit involving |x − 1| near 1 must be done *twice*, once from each side.

**Vertical asymptotes as one-sided limits:** if lim x→a⁺ f(x) = ∞ and lim x→a⁻ f(x) = ∞, write lim x→a f(x) = ∞. If one side is ∞ and the other −∞, the two-sided limit does not exist (but the vertical asymptote x = a still stands). Denominator → 0 with numerator → nonzero is the signature of an infinite limit: check the sign on each side.

**Watch out**

 −1/(x − 2)² blows up on both sides, but −1/(x − 2) blows up to +∞ on the left and −∞ on the right. Even-power denominators keep one sign; odd powers flip it. Check the sign, not just the blow-up.

## 6 · Poly-type functions and the algebra behind them

"Poly-type" limits wrap polynomials in roots, absolute values or compositions. The meta-strategy never changes: (1) try substitution; (2) if 0/0, factor or conjugate; (3) if ∞/∞, divide by the highest power; (4) if |·| is present, split into cases; (5) if stuck, check whether both sides or both infinities agree.

**Chapter formula sheet**

0/0 rational: factor and cancel, then substitute

Radical difference: multiply by conjugate A + B, use (A − B)(A + B) = A² − B²

∞/∞: divide every term by the highest power of x

deg p < deg q ⇒ 0 · deg p = deg q ⇒ a/b · deg p > deg q ⇒ ±∞

|x − a| = (x − a) if x ≥ a, and −(x − a) if x < a

√(x²) = |x| — minus sign applies when x → −∞

## Common mistakes

  - Canceling (x − 3) when x = 3 is exactly where the original is undefined without noting the hole — the shortcut is fine for limits, but say why it is legal.

  - Multiplying by the conjugate of only the numerator and not the denominator (or vice versa).

  - Writing √(x²) = x for negative x — it is |x|, and it flips sign at −∞.

  - Declaring an infinite limit exists because the absolute value blows up, while the signed sides go to opposite infinities.

  - Using the degree rule when the expression is a sum or difference, not a quotient — it applies to fractions only.

  - Forgetting that |x − a| limits need a left case and a right case.

## Practice exercises

1. lim x→2 (x² − 4)/(x − 2) equals:

- A. 0
- B. 2
- C. 4
- D. undefined

2. Find lim x→1 (x³ − 1)/(x − 1).

3. lim x→0 (√(x + 9) − 3)/x = ?

4. Compute lim x→∞ (6x² + 1)/(2x² − 5).

5. lim x→∞ (x³ + 2x)/(x² + 1) is:

- A. 0
- B. 1
- C. ∞ (unbounded)
- D. 2

6. lim x→−∞ (√(x² + 1))/(2x) equals:

- A. 1/2
- B. −1/2
- C. 0
- D. does not exist

7. lim x→0 |x|/x from the right (i.e. 0⁺) equals what number?

8. Compute lim x→∞ (3x + 2)/(x² − 4).

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