Class 12 Statistics Notes · GSEB

Working Rules

Working Rules — apply the sum, product, quotient and constant multiple rules of limits to break complex expressions into simpler parts. GSEB Class 12 Commerce Statistics notes with rule cards and worked examples.

Last updated: 23 Sep 2026

Notes

The Four Rules

Working Rules of Limits

Algebraic laws that break a complicated limit into simpler pieces — solve each piece, then recombine. This is how you find limits WITHOUT building a table every time.

Like splitting a group project into parts, estimating each, and adding up — never attack a compound limit without splitting it first.

Rule Explorer — pick a rule, see it applied

Formula

Rule 1 — Sum Rule

\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)

Plain-text form

lim(f + g) = lim f + lim g

Worked example

lim(x→2) (x² + 3x) = 4 + 6 = 10

Split: lim(x→2) x² + lim(x→2) 3x
Polynomials: plug in → 2² = 4 and 3(2) = 6
Add: 4 + 6 = 10

Condition: Both individual limits must exist.

Quick check before using any rule

Do the individual limits exist? (If not, stop.)
Quotient rule only: is the denominator's limit ≠ 0? (If 0, stop and factor first.)
Split, substitute, simplify — that is your limit.

Watch Out: When the Quotient Rule Fails

Two traps, one fix

Denominator → 0: e.g. lim(x→5) (x − 5)/(x² − 25). Rule 3 is illegal — factor first: 1/(x + 5) → 1/10.

Got 0/0: not a dead end — factorise, cancel the common term, then substitute. (Standard Forms page covers the shortcuts.)

Tricks: Find Limits Without a Table

The method ladder — run down it until something clicks. No table needed.

Polynomial? Plug in directly

Built from +, −, × and powers only → substitute x = a. Done.

lim(x→2)(3x² + 5) = 3(4) + 5 = 17

Got 0/0? Factor and cancel

Factorise numerator (and denominator), cancel the common factor, then plug in.

lim(x→1)(x² − 1)/(x − 1) → cancel (x − 1) → x + 1 → 2

Spot (xⁿ − aⁿ)/(x − a)? Write n·aⁿ⁻¹

The key standard form — answer in one line, no working. (Standard Forms page has the full list.)

lim(x→3)(x² − 9)/(x − 3) = 2 × 3 = 6

Big expression? Split with the 4 rules

Break into sums/products/quotients of simple pieces, solve each, recombine.

lim(x→2)(3x² + 5x − 4) → 12 + 10 − 4 = 18

Both sides disagree? No limit

Quick check: evaluate from left and right (substitute a ± tiny step). Different answers → limit does not exist.

LHL = 1, RHL = 3 → no limit (jump)

Combining the Rules — Worked Example

Find lim(x→2) [3x² + 5x − 4]

Step 1 — Sum rule (repeatedly): split into lim 3x² + lim 5x − lim 4

Step 2 — Constant multiple: pull out 3 and 5 → 3·lim x² + 5·lim x − lim 4

Step 3 — Substitute: 3·(4) + 5·(2) − 4

Step 4 — Compute: 12 + 10 − 4 = 18

Answer

3(2)^2 + 5(2) - 4 = 12 + 10 - 4 = 18

Find lim(x→1) [(x² − 1)/(x − 1)] — the 0/0 trap

Direct substitution: (1 − 1)/(1 − 1) = 0/0 → quotient rule blocked. Factor first!

Factor: (x − 1)(x + 1)/(x − 1) = x + 1 for x ≠ 1

Now substitute: lim(x→1)(x + 1) = 2

After Cancelling

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1) = 2

Solved Examples

Solved Example

Problem

Evaluate lim(x→3) (2x + 7) using the working rules.

Solution

Solved Example

Problem

Is the quotient rule valid for lim(x→5) (x − 5)/(x² − 25)? Check before applying.

Solution

Rule invalid as-is; after factoring the limit is 1/10

Solved Example

Problem

Find lim(x→0) (5 · sin x / x) if lim(x→0) sin x / x = 1.

Solution

5 × 1 = 5

Key Takeaways

Four rules split hard limits into easy pieces: Sum, Product, Quotient, Constant Multiple.
Individual limits must exist — and for quotients, the denominator must not tend to 0.
Got 0/0? Do not divide — factorise, cancel, then substitute.
No-table ladder: plug in → factor/cancel → spot n·aⁿ⁻¹ → split with rules → check both sides.
Different LHL and RHL means no limit — always glance at both sides before declaring an answer.