Limits pass through sums, products and quotients

What to assume they know

• Why what a function approaches need not be what it equals — a limit read off nearby values, independent of the value at the point
• Approaching from the left and from the right, and when the two disagree — the existence criterion for a limit
• Algebraic manipulation of sums, differences, products and quotients of expressions
• That division by zero produces no number
• Each standard function is pinned down by its picture as much as by its rule — combining functions by adding and multiplying them

What they should be able to do

• State each of the four parts of Theorem 1 in your own words, including the hypothesis that both limits exist
• Explain why the hypothesis is needed, by exhibiting two pieces whose own limits both fail at a point while their sum behaves perfectly. Note why both have to fail: if a sum has a limit and one piece has a limit, the other is forced to have one too, as the difference of the first two. A single failure is available for a product, not for a sum
• State the condition attached to the quotient part and say what goes wrong without it
• Derive the constant-multiple rule as the special case of the product part in which one function is constant
• Verify the sum and product parts numerically on the chapter's own two worked checks, at x → 1 and at x → 0
• Extend the sum part from two functions to finitely many, and say why that extension is legitimate
• Identify, in a multi-step limit computation, which part of Theorem 1 licenses each step
• Recognise a computation in which Theorem 1 has been applied illegitimately

Where it usually goes wrong

• "These rules are obvious, so the theorem is bookkeeping." They are not obvious enough for this book to prove them, and the chapter says so. A limit is a statement about the behaviour of infinitely many values; that two such statements recombine cleanly is a real fact about the real numbers.
• "If the combination has a limit, the pieces must have limits." False, and section 10's example shows it. The theorem runs one way only.
• "The quotient rule works whenever the denominator function is non-zero near the point." The condition is on the denominator's limit, not on its values. Example 2 (iii)'s denominator is non-zero at every x near 2 and other than 2, and the rule still does not apply, because its limit at 2 is 0.
• "0/0 means the answer is 0" or "means the answer is 1." It means the quotient rule is unavailable and something else must be done. §12.3.2 and §12.4 are that something else.
• "You can apply the sum rule to a hundred terms because it says so." It says two. Extending to finitely many is a short induction, which the chapter performs silently on p. 229 when it splits a polynomial term by term. Make the step visible.
• "A constant multiplier is a separate rule to memorise." It is the product rule with a constant function on one side, and the chapter prints it as a Note for exactly that reason.

Questions to check understanding

• Evaluate a limit of a sum, difference, product or quotient by naming which part of Theorem 1 each step uses
• State the condition on the quotient part and give a function at which it fails
• Prove or disprove: if the limit of f + g exists then both limits exist
• Pull a constant out of a limit and justify the step from the product part
• One-mark: state the hypothesis of Theorem 1
• Evaluate a limit that needs the constant-multiple form together with the sum form, such as Exercise 12.1 q9 and q11 (p. 237)

Examples worth working on the board

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

• Theorem 1 (§12.3.1, p. 228). The setting: two functions f and g, and a point a at which both of their limits are assumed to exist. Under that assumption the theorem asserts four things — that the limit of the sum is the sum of the two limits, the limit of the difference is their difference, the limit of the product is their product, and the limit of the quotient is the quotient of the two limits, this last one carrying the extra requirement that the denominator's limit is not zero. Printed with the explicit remark that no proof is given.
• The Note under Theorem 1 (p. 228). Taking g to be the constant function with value λ, the product part becomes: a fixed multiplier passes straight through a limit. Verified by derivation: the limit of a constant function is that constant, by Illustration 4 (p. 223), and substituting that into the product part gives exactly the Note's statement. Show the substitution; the Note is not a fifth rule.
• The check inside Illustration 5 (p. 224). At x → 1 the chapter asks the student to confirm three facts first — that x² has limit 1, that x has limit 1, and that x + 1 has limit 2 — and then to use them twice. Verified: sum route, 1 + 1 = 2, which agrees with the limit of x² + x read off Table 12.7. Product route, x(x + 1) is the same function, and 1 × 2 = 2. The point of printing both is that two different decompositions of one function give the same answer, which is what a rule has to do to be a rule.
• The check inside Illustration 7 (p. 225). At x → 0 the chapter asks whether the limit of x + cos x splits into the limit of x plus the limit of cos x. Verified: 0 + 1 = 1, agreeing with the value 1 read off Table 12.9. Worth doing because cos x is not a polynomial — the check shows the sum rule does not care what kind of function it is applied to, only that both limits exist.
• Example 2 (iii) (§12.3.2, pp. 230–231) — the quotient condition failing in public. The function is (x² − 4)/(x³ − 4x² + 4x) at x → 2. Verified: numerator factors as (x + 2)(x − 2), denominator as x(x − 2)², so after cancelling one factor of (x − 2) the function is (x + 2)/(x(x − 2)), whose numerator tends to 4 and whose denominator tends to 0. The chapter records this one as not defined. Use it as section 8: the quotient part of Theorem 1 is simply unavailable here, and no amount of algebra makes it available.
• The two-scenario analysis (§12.3.2, p. 229). Where a rational function's denominator vanishes at a, the chapter splits on whether the numerator also vanishes there. If it does not, the limit fails. If it does, both can be written with powers of (x − a) pulled out, and the comparison of those two powers decides the outcome.
• An unlicensed application, for section 10 (not in the book, not printed). Take f(x) = |x|/x and g(x) = −|x|/x at 0. Neither has a limit there, by Approaching from the left and from the right, and when the two disagree. Their sum is identically 0 away from 0, so the sum has limit 0. The sum rule would have given nothing here, because its hypothesis was never met — and the example shows the converse is false: a combination can have a limit when the pieces do not.

Figures to have open

• A single card holding Theorem 1's four parts stacked under one hypothesis bar, reused as an overlay whenever a later topic invokes it. Standard schematic.
• A side-by-side of two decompositions of x² + x, both terminating at 2. Standard schematic.
• A "condition failed" stamp graphic for the quotient clause, reusable in Substitution works until it gives nothing over nothing, and then cancellation does and Trapping a function between two others to settle the trigonometric cases.
• No textbook figure is required for this topic. §12.3.1 prints none.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 12 "Limits and Derivatives", §12.3.1 Algebra of limits, printed p. 228, including the Note beneath Theorem 1.
• The two numerical checks the chapter sets for the student are at the foot of Illustration 5, p. 224, and the foot of Illustration 7, p. 225.
• The first application of the theorem, and of its failure case, is §12.3.2, pp. 228–231, in particular Example 2 (iii) on p. 231.
• Chapter Summary, p. 254, which restates the sum, difference, product and quotient statements in compressed form.
• Exercise 12.1, pp. 237–239. Of its first twelve items, seven really are direct applications of this topic's theorem; the other five — items 6, 7, 8, 10 and 12 — come out 0/0 at the stated point and need the cancellation of §12.3.2, with item 10 also needing the rational-exponent extension. Do not present the run of twelve as a single uniform block.