Substitution works until it gives nothing over nothing, and then cancellation does

What to assume they know

• Limits pass through sums, products and quotients — Theorem 1 and its quotient condition
• Why what a function approaches need not be what it equals — that a limit ignores the value at the point
• Factorising quadratics and cubics, and recognising (x − a) as a factor when a root is known
• Adding two algebraic fractions over a common denominator
• Laws of exponents including fractional exponents
• A rule that turns a position number into a term — the binomial expansion, used only as background

What they should be able to do

• Derive the limit of xⁿ at a from the product part of Theorem 1 by induction on n
• Show that a polynomial's limit at a point is its value there, naming the rule that licenses each line of the derivation
• Evaluate the limit of a rational function whose denominator does not vanish
• Diagnose a 0/0 form, extract the common factor of (x − a) from numerator and denominator, and state why cancelling it is allowed
• Determine, after cancellation, whether the limit is a number or fails to exist
• Combine two algebraic fractions into one before applying any of the above
• State and prove Theorem 2, the limit of (xⁿ − aⁿ)/(x − a), by dividing out the factor
• Apply Theorem 2 to a quotient of two such expressions, and to a surd by a substitution that makes it fit the theorem's shape

Where it usually goes wrong

• "Limits are found by substituting; the tables were just a slow way of doing it." Substitution is a theorem about polynomials and about rational functions with living denominators, and it is false in general — the chapter has already shown three functions where it gives the wrong answer or no answer.
• "0/0 equals 1, or 0, or is an error." Example 2 produces 0, then an undefined case, then −4, then 2 from four expressions all of that form. The form carries no information at all; it is a signal to factorise.
• "You cannot cancel (x − a) because it might be zero." At the one point where it is zero, the limit does not look. Everywhere the limit does look, (x − a) is a non-zero number and cancelling is ordinary arithmetic.
• "If the denominator tends to 0, the answer is infinity." Example 2 (iii) is recorded by the chapter as not defined, not as infinity. This book sets unbounded limits aside as outside the course (p. 226).
• "After one cancellation you are always done." In Example 2 (ii) and (iii) a squared factor of (x − 2) sits on one side of the fraction or the other — on top in (ii), underneath in (iii). One cancellation leaves another factor behind, and which of the two functions you are looking at decides whether the leftover sits on top or underneath.
• "Theorem 2 is a separate formula to memorise." Its proof is one factorisation and one substitution, and it is worth deriving every time rather than quoting.
• "Theorem 2 needs n to be a whole number." The theorem is stated for positive integers, and its Remark extends it to rational exponents with a positive. Example 3 (ii) and Exercise 12.1 q10 both depend on the extension.

Questions to check understanding

• Evaluate a polynomial limit by substitution and name the theorem used
• Evaluate a rational limit that is 0/0, showing the factorisation and stating the restriction under which the cancellation holds
• Decide whether a given rational limit exists, and if not, say why
• Combine two algebraic fractions and then evaluate the limit of the result
• Apply Theorem 2 to a quotient of two power-difference expressions, including fractional exponents
• Use a substitution to convert a surd expression into Theorem 2's shape
• Given a condition such as a + b + c ≠ 0 attached to a question, explain what it is protecting

Examples worth working on the board

Values marked verified are worked out here from the chapter's printed data; no answer key was consulted.

• The induction for xⁿ (§12.3.2, pp. 228–229). Base: the limit of x at a is a. Step: the limit of x² is got by writing x² as x times x and applying the product part of Theorem 1, giving a·a. The chapter then says an induction on the exponent settles every n, and records the limit of xⁿ at a as aⁿ. Verified as a derivation: if the limit of xᵏ is aᵏ, then xᵏ⁺¹ = xᵏ·x has limit aᵏ·a = aᵏ⁺¹ by the same product part. Show that one line; the chapter compresses it to a phrase.
• Polynomials (p. 229). With f(x) = a₀ + a₁x + a₂x² + … + aₙxⁿ, the chapter splits the limit across the terms by the sum part, pulls each coefficient out by the Note's constant rule, replaces each limit of a power by the corresponding power of a, and reassembles — arriving at f(a). The printed text asks the reader to justify every step; that instruction is the section.
• Rational functions (p. 229). f = g/h with g and h polynomials. Where h(a) is not zero, the quotient part gives g(a)/h(a) directly. Where h(a) = 0 there are two cases: if g(a) is not zero the limit fails; if g(a) = 0 as well, write g(x) = (x − a)ᵏ g₁(x) and h(x) = (x − a)ˡ h₁(x), taking k as the largest power of (x − a) dividing g. The chapter then treats k > l, which gives 0, and states that for k < l the limit is not defined. Both inequality directions read off pp. 229–230.
• Example 1 (p. 230), three polynomial limits, all by substitution: x³ − x² + 1 at x → 1; x(x + 1) at x → 3; and 1 + x + x² + … + x¹⁰ at x → −1. Verified: 1 − 1 + 1 = 1; 3 × 4 = 12; and the third has eleven terms with signs alternating from +1, so the eleven terms cancel in pairs and leave 1.
• Example 2 (pp. 230–232), five rational limits. The data, as printed: (i) (x² + 1)/(x + 100) at x → 1; (ii) (x³ − 4x² + 4x)/(x² − 4) at x → 2; (iii) (x² − 4)/(x³ − 4x² + 4x) at x → 2; (iv) (x³ − 2x²)/(x² − 5x + 6) at x → 2; (v) (x − 2)/(x² − x) − 1/(x³ − 3x² + 2x) at x → 1. Verified: (i) substitution gives 2/101. (ii) 0/0; factor to x(x − 2)²/((x + 2)(x − 2)), cancel one (x − 2), giving x(x − 2)/(x + 2) whose limit is 0. (iii) 0/0; the same two factorisations the other way up give (x + 2)/(x(x − 2)), which tends to 4/0 — not defined. (iv) 0/0; factor to x²(x − 2)/((x − 2)(x − 3)), cancel, giving x²/(x − 3) whose limit is 4/(−1) = −4. (v) combine the two fractions over x(x − 1)(x − 2), giving ((x − 2)² − 1)/(x(x − 1)(x − 2)) = (x² − 4x + 3)/(x(x − 1)(x − 2)) = (x − 3)(x − 1)/(x(x − 1)(x − 2)); cancel (x − 1) and substitute, giving (1 − 3)/(1 × (1 − 2)) = 2. Note how differently the four 0/0 cases end — 0, undefined, a negative number, and a positive number. That spread is the reason to work all of them.
• The restriction printed with the cancellation (pp. 231–232). Each cancellation in Example 2 is annotated with the fact that the variable is not equal to the point. That annotation is the justification, not a caution.
• Theorem 2 (p. 232). For a positive integer n, the limit of (xⁿ − aⁿ)/(x − a) at x → a is n·aⁿ⁻¹. Proof (p. 233): dividing xⁿ − aⁿ by x − a leaves a sum whose terms run xⁿ⁻¹, xⁿ⁻²a, …, aⁿ⁻¹; substituting x = a into that sum gives n copies of aⁿ⁻¹. Verified: the sum has exactly n terms, the exponents on x and on a in each term adding to n − 1.
• The Remark under Theorem 2 (p. 232). The formula survives when n is any rational number, provided a is positive. Not proved here.
• Example 3 (p. 233). (i) (x¹⁵ − 1)/(x¹⁰ − 1) at x → 1. Verified: divide top and bottom by (x − 1) to make two copies of Theorem 2's shape with a = 1, giving 15/10 = 3/2. (ii) (√(1 + x) − 1)/x at x → 0, handled by putting y = 1 + x so that y → 1 and the expression becomes (y^(1/2) − 1^(1/2))/(y − 1) — Theorem 2's shape with the rational exponent 1/2. Verified: the value is (1/2)·1^(−1/2) = 1/2.
• The recipe (end of §12.4, p. 237). If f(a) and g(a) are both zero, look for a factor of f that vanishes at a with a second factor that does not, do the same for g, cancel what is common, and substitute into what is left. This is printed after the trigonometric section but it governs everything in §12.3.2 as well.
• Exercise 12.1 items this topic owns (pp. 237–239): 1 to 12 and 27, 29 and 31. Item 10 is (z^(1/3) − 1)/(z^(1/6) − 1) at z → 1, which is Theorem 2 with rational exponents in both parts. Item 11 is (ax² + bx + c)/(cx² + bx + a) at x → 1 with a + b + c not zero — the condition is there precisely to keep the denominator's limit alive. Item 29 is the product (x − a₁)(x − a₂)…(x − aₙ), asked at a₁ and at a general point.

Figures to have open

• A decision tree: evaluate top and bottom at a → three branches (denominator alive; denominator dead and top alive; both dead) → the action each branch takes. This is the brief's organising picture and the chapter has no figure of its own here — §12.3.2 prints none, which I confirmed on pp. 228, 229 and 230.
• The long-division layout of xⁿ − aⁿ by x − a with the n terms of the quotient laid out and counted. Standard schematic.
• A four-panel result board for Example 2 (ii) to (v), showing that identical starting forms end at unlike answers.
• No textbook figure is needed for this topic.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 12 "Limits and Derivatives", §12.3.2 Limits of polynomials and rational functions, printed pp. 228–233, and the general recipe printed at the close of §12.4, p. 237.
• Example 1 (p. 230), Example 2 (pp. 230–232), Theorem 2 with its Remark (p. 232), the proof of Theorem 2 and Example 3 (p. 233).
• Exercise 12.1 items 1–12, 27, 29, 31, printed pp. 237–239.
• Chapter Summary, p. 254, which carries Theorem 2's formula among the standard limits.
• Backward reference inside the chapter: the cancellation used throughout this section is the one the falling-body discussion of §12.2 stopped short of, and the function of p. 220 is its first appearance.