Calculus I › Limits › full formula sheet

lim(x→a) [f(x) ± g(x)] = L ± M

The sum and difference laws

Break a limit of a sum into a sum of limits -- but only when every piece converges.

Notation: L and M are the two individual limits, assumed to exist and be finite.

Before this lesson: The constant multiple law

Notation in this lesson

L, M
the two individual limits — both must exist and be finite
ε/2
half the error budget: each piece gets ε/2 so the total is ε
δ = min(δ1, δ2)
the smaller neighborhood — where both approximations hold at once

Where it comes from

The problem these laws solve is divide and conquer. A limit like lim(x→2)(x² + 3x) looks tangled — but if you already know lim(x→2) x² = 4 and lim(x→2) 3x = 6, the answer should just be 4 + 6 = 10. The sum law is the license to split first and add later:

Before reading on: limx→2(x²+3x) — why should splitting it into two separate limits be legal? What's the license?
lim(x→a) [f(x) + g(x)] = lim(x→a) f(x) + lim(x→a) g(x)provided both limits on the right existSay it: the limit as x approaches a of f of x plus g of x equals the limit as x approaches a of f of x plus the limit as x approaches a of g of x

Before reading on: limx→0(1/x − 1/x) — split first, or simplify first? Predict the honest answer before you read on.
The tempting — and wrong — move is splitting before checking that the pieces converge. Watch it die:

lim(x→0) (1/x − 1/x)
=
lim(x→0) 0 = 0
The honest answer. For every x ≠ 0 the expression is exactly 0, so the limit is 0.
“split first”
=
lim(x→0) 1/x − lim(x→0) 1/x
Meaningless. Neither piece converges (one-sided: plus/minus infinity), so the law’s hypothesis fails and “infinity minus infinity” is not a number. Splitting created the mess; simplifying first avoids it.

So the law carries a hypothesis: both individual limits must exist (and be finite). Here is the intuition for why it works when they do. If f(x) stays within 0.01 of L and g(x) stays within 0.01 of M, then f(x) + g(x) stays within 0.01 + 0.01 = 0.02 of L + M. Errors add; they never multiply or explode. The proof below just makes “plus/minus 0.01” rigorous with half-epsilon.

The difference law needs no separate proof: f − g is f + (−g), and the constant multiple law gives lim(−g) = −M. One law, one sign flip.

Derivation

Assume lim(x→a) f(x) = L and lim(x→a) g(x) = M, both finite. We prove lim(x→a) [f(x) + g(x)] = L + M from the epsilon-delta definition.

|(f+g) − (L+M)|
=
|(f − L) + (g − M)| ≤ |f − L| + |g − M|
Step 1 — the triangle inequality. Regroup the total error as (error in f) + (error in g). The triangle inequality says the total error is at most the sum of the two errors. This is the whole proof in one line.
|f − L| < ε/2
and
|g − M| < ε/2
Step 2 — split the error budget. Since lim f = L, some δ1 forces |f − L| < ε/2; since lim g = M, some δ2 forces |g − M| < ε/2. Each piece gets half the budget — that is why ε/2 appears.
δ = min(δ1, δ2)
⇒
|(f+g) − (L+M)| < ε/2 + ε/2 = ε
Step 3 — take the smaller δ. With 0 < |x − a| < δ, both approximations hold at once (a smaller neighborhood satisfies both δ1 and δ2). The two half-errors add to ε. ∎

Key steps shown; the argument above is complete. The difference law follows immediately: lim(f − g) = lim(f + (−g)) = L + (−M) = L − M, using the sum law plus the constant multiple law (see its page). Note the dependency order: constant multiple first, then difference.

How to use it

The procedure, every time:

  1. Check the hypothesis. Do lim f and lim g both exist (finite)? If either diverges or oscillates, stop — the law does not apply. Simplify or combine first.
  2. Evaluate each piece separately, using whatever tool fits each one (direct substitution, factoring, another limit law).
  3. Add or subtract the results. Three or more terms? Apply the law twice: lim(f + g + h) = lim((f+g) + h).
  4. Sanity-check the sign. For differences, a surprising negative usually means a dropped minus, not a wrong law.

Split first, or simplify first?

If every piece converges, split immediately — it is the fastest path. If any piece blows up (infinity, oscillation), simplify the combined expression first (cancel, factor, combine fractions), then split. The rule of thumb: split what converges; combine what diverges.

Common mistake: writing lim(f + g) = lim f + lim g when one piece diverges, then “evaluating” infinity minus infinity. The law’s hypothesis is doing real work — check it before you split.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: lim(x→2) (x² + 3x)

  1. Check the hypothesis. lim(x→2) x² = 4 and lim(x→2) 3x = 6 both exist — safe to split.
  2. Evaluate each piece. lim x² = 4 (power law); lim 3x = 3·2 = 6 (constant multiple law).
  3. Add. 4 + 6 = 10.
  4. Check by direct substitution. 2² + 3·2 = 4 + 6 = 10. Matches ✓
Common mistake: none here — but notice the check in step 1. Skipping it is exactly what breaks Example 3.
Your turn: limx→3 (x²+2x) = ?

Answer: 15.

Both pieces converge: lim x² = 9 and lim 2x = 6, so 9 + 6 = 15.

Example 2 — with a root: lim(x→1) (√x − 1/x)

  1. Check the hypothesis. lim(x→1) √x = 1 and lim(x→1) 1/x = 1 both exist (x = 1 is safely inside both domains).
  2. Evaluate each piece. √1 = 1; 1/1 = 1.
  3. Subtract. 1 − 1 = 0.
Common mistake: combining into a single fraction first. Unnecessary — when both pieces converge, splitting is the fast lane.
Your turn: limx→4 (√x + 1/x) = ?

Answer: 9/4.

lim √x = 2 and lim 1/x = 1/4 both exist: 2 + 1/4 = 9/4.

Example 3 — the trap: lim(x→0) (1/x − 1/x)

  1. Check the hypothesis. lim(x→0) 1/x does not exist (one-sided: +infinity vs −infinity). Do not split.
  2. Simplify the combined expression first. For every x ≠ 0, 1/x − 1/x = 0 exactly.
  3. Now the limit is trivial: lim(x→0) 0 = 0.
Common mistake: splitting anyway and writing “infinity minus infinity = 0”. The answer 0 is right but the reasoning is illegal — infinity minus infinity is indeterminate, and on another problem the same move gives a wrong answer.
Your turn: limx→0 (1/x² − 1/x²) = ?

Answer: 0.

Don't split — both pieces diverge. Simplify first: 1/x² − 1/x² = 0 for every x ≠ 0.

Example 4 — three terms: lim(x→4) (√x + x³ − 2x)

  1. Check the hypothesis. All three pieces converge at x = 4 (√x is fine near 4).
  2. Apply the law twice. lim((√x + x³) − 2x) = lim(√x + x³) − lim 2x = (lim √x + lim x³) − lim 2x.
  3. Evaluate: 2 + 64 − 8 = 58.
Common mistake: arithmetic slips in multi-term problems. Write each piece’s limit down before combining.
Your turn: limx→9 (√x + x² − 3x) = ?

Answer: 57.

Three pieces, all convergent: apply the law twice → 3 + 81 − 27 = 57.

Memorization tips

  • Chant the hypothesis: “split only what converges.” Both pieces must have (finite) limits, or the law is silent.
  • The half-epsilon memory hook: two pieces, half the error budget each. If you ever write a limit proof, the budget split is the move to remember.
  • Difference = sum with a sign flip: f − g is f + (−g). One law covers both — never memorize them as separate facts.
  • Direct substitution on polynomials is this law in disguise: every polynomial is a sum of constant-multiple power terms, each continuous. “Plug in” is just the sum law on autopilot.
  • When pieces blow up, combine first, split later: simplify the whole expression (cancel, factor), then check whether the new pieces converge.
  • The min-delta trick generalizes: whenever two approximations must hold at once, take the smaller neighborhood. You will see it again in the product law proof.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and this law is yours.

← Back to the Calculus I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the sum law for limits?

If lim(x→a) f(x) = L and lim(x→a) g(x) = M both exist and are finite, then lim(x→a) [f(x)+g(x)] = L+M: the limit of a sum is the sum of the limits.

Why can't I always split a limit into pieces?

The law requires each piece's limit to exist. In lim(x→0)(1/x − 1/x), neither piece converges, so splitting gives the meaningless ∞−∞ — while simplifying first gives the correct answer, 0.

Why does the proof use ε/2?

The triangle inequality bounds the total error by the sum of the two pieces' errors. Giving each piece half the budget (ε/2) makes the total come out to exactly ε.

How is the difference law related to the sum law?

It is the sum law in disguise: f−g = f+(−g), and the constant multiple law gives lim(−g) = −M. So lim(f−g) = L−M with no new proof needed.

Do the sum and difference laws work for three or more functions?

Yes — apply the law repeatedly: lim(f+g+h) = lim((f+g)+h) = L+M+N, as long as every piece's limit exists.

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