Skill: limits

Limits

Find what functions approach: direct substitution, factoring, conjugates, the sin(x)/x limit, one-sided limits, and continuity.

1 What a limit is

A limit asks: what value is the function approaching as x approaches x₀? It never asks what happens at x₀.

The idea

lim_x→x₀ f(x) = L means f(x) gets arbitrarily close to L as x gets close to x₀

The function may have a hole at x₀ (or a completely different value there) and the limit still exists — the limit only cares about the approach, never the landing.

Example f(x) = (x² − 1)/(x − 1) at x = 1

Direct substitution gives 0/0 — a hole at x = 1. But for x ≠ 1, f(x) = x + 1, so values approach 2.

2 Algebraic methods

When substitution gives 0/0, algebra removes the hole first.

The toolkit

Direct: plug in x₀ — if defined, that is the limit
Factor: (x² − k²)/(x − k) = x + k, so the limit at k is 2k
Conjugate: (√(x+a) − b)/(x − c) → multiply by the conjugate; limit is 1/(2b)
Special: lim_x→0 sin(kx)/(mx) = k/m

Order of attack: try substitution → factor → conjugate → special limits.

Example lim_x→4 (√(x + 5) − 3)/(x − 4)

Substitution gives 0/0. Multiply top and bottom by (√(x+5) + 3): the top becomes (x − 4), which cancels, leaving 1/(√(x+5) + 3) → 1/6.

3 One-sided limits and continuity

A two-sided limit exists only when the left and right approaches agree.

The agreement test

lim_x→x₀ f(x) exists ⇔ lim_x→x₀⁻ = lim_x→x₀⁺

f is continuous at x₀ when three things hold: f(x₀) is defined, the limit exists, and they are equal. A jump discontinuity fails the agreement test.

Example f(x) = { 2x + 1, x < 1; 3x − 2, x > 1 } at x = 1

Left: 2(1) + 1 = 3. Right: 3(1) − 2 = 1. They disagree, so the two-sided limit does not exist — a jump.

4 Worked examples

One of each method, each verified independently.

Example 1 Direct substitution

Evaluate lim_x→0 (2x + 1)/(x + 1).

Denominator is 1 ≠ 0, so substitute: (0 + 1)/(0 + 1) = 1.

Example 2 Factoring

Evaluate lim_x→3 (x² − 9)/(x − 3).

0/0 → factor: (x − 3)(x + 3)/(x − 3) = x + 3 → 6.

Example 3 Conjugate

Evaluate lim_x→4 (√(x + 5) − 3)/(x − 4).

Conjugate: the fraction becomes 1/(√(x+5) + 3) → 1/6.

Example 4 The sine limit

Evaluate lim_x→0 sin(3x)/(5x).

k/m = 3/5.

Example 5 One-sided (left)

f(x) = { 2x + 1, x < 1; 3x − 2, x > 1 }. Find lim_x→1⁻ f(x).

Use the left piece: 2(1) + 1 = 3.

Example 6 One-sided (right)

Same f. Find lim_x→1⁺ f(x).

Use the right piece: 3(1) − 2 = 1.

5 Common mistakes

Three errors that break limit problems.

Answering 0/0 as the limit

Wrong: lim_x→3 (x² − 9)/(x − 3) = 0/0, so the limit is 0.
Right: 0/0 is not an answer — it is a signal to do algebra. Factor first: x + 3 → 6.

Forgetting to check both sides

Wrong: the left limit is 3, so the limit is 3.
Right: the two-sided limit exists only if both one-sided limits agree. Here the right limit is 1, so the two-sided limit DNE.

Confusing the limit with f(x₀)

Wrong: f has a hole at x₀, so the limit does not exist.
Right: the limit cares about the approach, never the landing. A hole at x₀ is exactly where limits do their best work.

6 Key vocabulary

Words to know

Indeterminate form — 0/0: not an answer, a signal to simplify.
Removable discontinuity — a hole; the limit exists but f(x₀) is missing or different.
Jump discontinuity — left and right limits disagree; the two-sided limit DNE.
Continuous at x₀ — f(x₀) defined, the limit exists, and they are equal.

7 Quick check

Try these before moving on — click to reveal each answer.

Evaluate lim_x→2 (x² − 4)/(x − 2).

Factor: x + 2 → 4.

Evaluate lim_x→0 sin(7x)/(3x).

k/m = 7/3.

f(x) = { x + 1, x < 0; x − 1, x > 0 }. Does lim_x→0 f(x) exist? Is f continuous at 0?

Left = 1, right = −1 — they disagree, so the limit DNE and f is not continuous at 0.

8 Next steps

Now drill the skill with endless randomized problems.

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