Calculus I › Limits › full formula sheet

limx→a ⁿ√f(x) = ⁿ√L

Say it: “the limit as x approaches a of the nth root of f of x equals the nth root of L”

The root law for limits

Take the root of the limit — after checking the root is allowed to exist.

On this page L = limx→a f(x) exists (finite). For even n we need L > 0 (and f ≥ 0 near a); for odd n any real L works. These conditions are the whole page.

Before this lesson: The quotient law

Notation in this lesson

n√
the nth root
L
the inner limit — check it before taking the root
even n: L > 0
even roots need a positive inside; odd roots accept any real L

Where it comes from

The root law is the power law in reverse: roots undo powers, so limits should pass through them the same way. And they do — with domain conditions attached, because roots are pickier than powers:

Before reading on: limx→−4 √x — the inside → −4. Can you just take the root of the limit? What's the catch?
lim(x→−4) √x
=
“√(−4)” = ???
Broken. The inner limit is −4, but √(−4) isn’t a real number — √x isn’t even defined on any punctured neighborhood of −4. No two-sided real limit exists. The “L > 0 for even n” condition exists precisely to forbid this.
lim(x→−8) ∛x
=
∛(−8) = −2
Fine. Odd roots accept negative inputs, so no extra condition is needed. Parity is everything.

The intuition: roots are continuous wherever they are defined, so “limit then root” equals “root then limit.” The derivation makes this precise with the conjugate trick — the same rationalizing move you will use on 0/0 root problems forever.

Derivation

We prove the square-root case (n = 2); the pattern generalizes. Assume lim(x→a) f(x) = L with L > 0 (so √f is defined near a). Goal: √f(x) → √L.

√f − √L
=
(f − L) / (√f + √L)
Step 1 — the conjugate trick. Multiply top and bottom by (√f + √L): the numerator becomes (√f)² − (√L)² = f − L. The difference of roots becomes a difference of insides over a sum of roots.
numerator → 0
and
denominator → 2√L > 0
Step 2 — evaluate the pieces. The numerator f − L → 0 is exactly the hypothesis. The denominator → √L + √L = 2√L, which is positive — this is where L > 0 earns its keep.
lim(√f − √L)
=
0 / (2√L) = 0
Step 3 — quotient law. The denominator stays away from zero, so the quotient law applies: 0 divided by anything nonzero is 0. Hence √f → √L. ∎

Key steps shown; the argument above is complete for n = 2. For general n, the same idea uses An − Bn = (A−B)(An−1 + An−2B + … + Bn−1), giving ⁿ√f − ⁿ√L = (f−L)/(a sum of n terms → n·L(n−1)/n > 0). For odd n the denominator stays nonzero even at L = 0, which is why odd roots need no positivity condition. At L = 0 with even n, use one-sided limits (e.g. lim(x→0+) √x = 0).

How to use it

The procedure, every time:

  1. Check parity and the inner limit. Even n: need L > 0 strictly (L = 0 allows only one-sided). Odd n: any real L is fine.
  2. Check the domain near a: is ⁿ√f(x) actually defined on a punctured neighborhood (or the relevant side)? If not, the two-sided limit DNE.
  3. Evaluate the inner limit L, then take the root: ⁿ√L.
  4. Combine freely with sum/product/quotient laws for messier expressions.

The L = 0 edge case

For even n with inner limit 0, the two-sided law technically needs f ≥ 0 on both sides. In practice you meet this as a one-sided limit: lim(x→0+) √x = 0 is perfectly fine. Name the side and move on.

Common mistake: “√(−9) = −3”. No — over the reals, even roots of negatives don’t exist. If your inner limit comes out negative under an even root, the answer is DNE (reals), not a negative number.

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→9) √x

  1. Check conditions. Even root (n = 2); inner limit 9 > 0. √x is defined near 9. All good.
  2. Take the root: √9 = 3.
Common mistake: “±3”. The radical symbol √ means the principal (nonnegative) root. ± appears when solving x² = 9, not when evaluating √9.
Your turn: limx→16 √x = ?

Answer: 4.

Even root, inner limit 16 > 0: √16 = 4.

Example 2 — odd root, negative inside: lim(x→2) ∛(8x−8)

  1. Check conditions. Odd root (n = 3) — any real inner limit works.
  2. Inner limit: 8·2 − 8 = 8.
  3. Take the root: ∛8 = 2.
Common mistake: worrying about the sign of the inside for odd roots. Don’t — cube roots eat negatives happily.
Your turn: limx→1 ∛(9x−1) = ?

Answer: 2.

Odd root — any inner limit works: 9·1 − 1 = 8, and ∛8 = 2.

Example 3 — root of a sum: lim(x→5) √(2x−1)

  1. Check conditions. Even root; inner limit 2·5 − 1 = 9 > 0. Fine.
  2. Inner limit first (sum + constant multiple laws): 9.
  3. Take the root: √9 = 3.
Common mistake: “distributing” the root: √(2x) − √1. Roots don’t distribute over sums — evaluate the inside first.
Your turn: limx→4 √(5x−4) = ?

Answer: 4.

Inner: 20 − 4 = 16 > 0, so √16 = 4.

Before reading on: √x near −9 — even root, negative inside. Predict: what breaks?

Example 4 — the trap: lim(x→−9) √x

  1. Check conditions. Even root; inner limit −9 < 0. Condition fails.
  2. Check the domain: √x is undefined for all x < 0, so no punctured neighborhood of −9 works.
  3. Verdict: the (two-sided, real) limit DNE.
Common mistake: writing √(−9) = −3 or 3i. Over the reals — this course’s universe — the expression is simply undefined.
Your turn: limx→−16 √x = ?

Answer: DNE.

Even root with inner limit −16 < 0 — √x isn't defined on any punctured neighborhood of −16.

Memorization tips

  • Parity first: even root → demand L > 0; odd root → relax. Ask “even or odd?” before anything else.
  • The conjugate trick is the takeaway: √f − √L = (f−L)/(√f+√L). You will reuse this rationalization on 0/0 root problems for the rest of the course.
  • √ means principal root: √9 = 3, never ±3. The ± belongs to solving equations, not evaluating radicals.
  • Roots don’t distribute: √(a+b) ≠ √a + √b. Inside first, root last — always.
  • L = 0 under an even root → think one-sided: lim(x→0+) √x = 0 is fine; the two-sided version needs the domain on both sides.
  • Root law + power law = inverse pair: they undo each other, which is why their proofs mirror each other (induction one way, conjugates the other).

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 root law for limits?

If lim(x→a) f(x) = L exists, then lim(x→a) ⁿ√f(x) = ⁿ√L — with conditions: for even n you need L > 0 (and f ≥ 0 near a); for odd n any real L works.

Why do even roots need L > 0?

Even roots of negative numbers aren't real, so ⁿ√f(x) wouldn't even be defined near a. The proof also needs the conjugate-sum denominator to stay away from zero, which L > 0 guarantees.

What is the conjugate trick in the proof?

√f − √L = (f−L)/(√f+√L): multiplying by the conjugate turns a difference of roots into a difference of insides over a sum of roots, so the quotient law finishes the job.

Is √9 = ±3?

No — the radical symbol √ means the principal (nonnegative) root, so √9 = 3. The ± appears when solving x² = 9, not when evaluating √9.

What happens when the inner limit is 0 under an even root?

Use one-sided limits: lim(x→0⁺) √x = 0 is fine. The two-sided limit needs the function defined on both sides of the point.

More from the codex

Live

Calculus I formula sheet

All 71 formulas — limits, derivatives, integrals — printable and quiz-ready.

Open sheet → Live

Formula Sheet Builder

Mix and match any sections into your own printable sheet.

Open tool → Live

Prompt Simulator

Practice prompt engineering with deterministic scoring.

Open tool →