Combining continuous functions, and why a composite survives too

What to assume they know

• Definition 1 and Definition 2, and continuity as an agreement between limit and value
• The algebra of limits from Class XI, including the quotient case and its proviso
• Composition of two functions, and the condition on ranges that makes it defined
• Polynomial and rational functions, and where a rational function is undefined
• The addition formula for the sine of a sum, from Class XI
• The tangent, cosecant, secant and cotangent as quotients of sine and cosine
• The modulus function and its two-branch form

What they should be able to do

• State the four parts of Theorem 1 and identify which one carries a proviso
• Reproduce the proof of the sum part and say at which line continuity of the two functions is used
• Adapt that proof to the difference, the product and the quotient
• Apply the two Remarks to a constant multiple and to a reciprocal
• Deduce that a quotient of two polynomials is unbroken across its whole domain
• Reproduce the chapter's argument for the sine and name every limit it consumes
• Deduce continuity of the cosine, the tangent and the three reciprocal ratios, and state the excluded inputs for each
• State Theorem 2 precisely, including which function must be continuous at which point
• Decide continuity of a composite by naming an inner and an outer function
• Recognise a function that Theorem 1 and Theorem 2 together do not settle, and say why

Where it usually goes wrong

• "Theorem 1 is four separate facts to memorise." It is one substitution run four times. The chapter proves one and calls the rest similar, which is the strongest possible hint about how to revise them.
• "The quotient part needs the divisor to be non-zero everywhere." It needs it at the one input under discussion. That is why the theorem still applies to the tangent at every input where the tangent exists.
• "Theorem 2 needs both functions continuous at the same point." It does not. The outer function is asked for continuity at the value the inner function produces. Say the two places out loud when applying it.
• "A composite of continuous functions is continuous, full stop." Only where the composite is defined and the inner function is continuous. Exercise 5.1 Q24 is the counter-case: the inner function has no value at zero and the theorem is silent there.
• "The chapter proved the sine is continuous." It gave an argument resting on one limit it declined to prove and a second limit it never mentioned. That is still a good argument; it is not a proof from first principles, and a student should know which it is.
• "Every trigonometric function is continuous everywhere." Four of the six lose infinitely many inputs, and the two excluded sets are different. Naming which set goes with which function is a routine examination question.
• "Continuity of a product needs both factors non-zero." Nothing of the kind. Only division carries a proviso, and only about the divisor.
• "If Theorem 1 and Theorem 2 do not apply, the function is discontinuous." They are sufficient conditions, not a test. Exercise 5.1 Q24 is continuous at the very input where neither theorem reaches.

Questions to check understanding

• State Theorem 1 and prove one part other than the sum
• Justify continuity of a stated rational function and give its domain
• Decide continuity of a sum, difference or product of two named trigonometric functions — the form of Exercise 5.1 Q21
• Name the excluded inputs for the cosecant, the secant and the cotangent, and justify each from Theorem 1 — the form of Exercise 5.1 Q22
• Split a given function into a named inner and a named outer function, then apply Theorem 2 — the form of Exercise 5.1 Q31 to Q33
• Decide continuity at a point at which a stated piecewise function has been given a value chosen to match its limit — the form of Exercise 5.1 Q25
• Identify an input at which Theorem 1 and Theorem 2 do not settle the question, and say what extra argument is needed

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• The opening paragraph of §5.2.1 (Part I p. 113). It says three things in order: limits were given an algebra in the previous class; continuity is decided entirely by a limit; therefore matching results should be expected here. That is the argument of section 1, and it is an argument about why the results are unsurprising, not a proof of any of them.
• Theorem 1 (Part I p. 113). Four parts — sum, difference, product, quotient — each asserting continuity at the same input, with the fourth carrying the condition that the divisor does not vanish there. Note the shape of the hypothesis: both functions are assumed continuous at one particular input, not on an interval. That is why the theorem applies at a single join as easily as everywhere at once.
• The proof of the sum (Part I p. 113). Four lines, each annotated in the margin with its justification: the definition of a sum of functions; the limit theorem; the continuity of the two functions; the definition of a sum again. Verified: continuity of the two functions is used once, in the third line, and nowhere else. The other three lines are bookkeeping. Show the annotations — they are the skeleton every one of the remaining three proofs shares.
• The sentence that closes the proof (Part I p. 113). The remaining three parts are described as similar and handed to the reader. Verified by writing them out: the difference is character-for-character the same with one sign changed; the product replaces the sum limit theorem with the product one; the quotient does the same and needs the divisor's limit at the input to be non-zero, which is exactly where the printed proviso comes from.
• The two Remarks (Part I p. 114). The first specialises the product part to a constant function, giving a constant multiple, and then to minus one, giving the negative of a continuous function. The second specialises the quotient part to a constant numerator, giving the reciprocal wherever the divisor does not vanish. Verified: both are one substitution each. They matter because almost every function in Exercise 5.1 is assembled from these two moves plus the four parts.
• Example 16 (Part I p. 114). Every rational function is continuous, in four lines: it is a quotient of polynomials, polynomials are continuous by the previous topic's Example 14, and the fourth part of Theorem 1 does the rest. Verified: the domain has to exclude the zeros of the divisor, which the chapter states, and on that domain the proviso holds at every input by construction — so the exclusion is not an extra hypothesis, it is the domain.
• Example 17 (Part I p. 114). Continuity of the sine. The chapter lists one fact — that the sine tends to zero at zero — writes the input as a fixed number plus a small increment, expands with the addition formula, splits the limit in two, and lands on the sine of the fixed number. The finding to show: the argument uses a second limit that is never listed. Verified against the printed chain: the first of the two split terms collapses to the sine of the fixed number only if the cosine of the increment tends to one, and that limit appears nowhere on the page. One of the two limits the proof consumes is declared; the other is silent. This is worth thirty seconds, and it is not a complaint — it tells a student which line of their own version needs a sentence.
• The Remark after Example 17 (Part I p. 115). Continuity of the cosine is delegated with one sentence saying a similar proof works. Verified: it does — the addition formula for the cosine of a sum plus the same two limits.
• Example 18 (Part I p. 115). The tangent, as a quotient of two functions now known continuous, is unbroken at every input where it exists; the chapter names the excluded inputs as the odd multiples of half pi. Verified: those are exactly the zeros of the cosine, and the proviso of Theorem 1's fourth part fails at precisely those inputs and nowhere else.
• Theorem 2 (Part I p. 115). The composite is continuous at an input provided the inner function is continuous at that input and the outer function is continuous at the inner function's value there. The chapter states it without proof. The asymmetry is the content: the two functions are not asked for continuity at the same place, and a student who reads it as "both continuous at the same input" will misapply it constantly.
• Examples 19 and 20 (Part I p. 115). A sine of a square, and a modulus of an expression built from a linear term and a modulus. Verified: the first puts the squaring rule inside the sine, and both are unbroken everywhere, so Theorem 2 applies at every input. The second needs Theorem 1 first — the inner function is a polynomial plus a modulus, continuous as a sum — and then Theorem 2 with the modulus as the outer function. Example 20 is the better one, because it uses both theorems in one line and the chapter's own sentence ordering makes that easy to miss.
• Exercise 5.1 Q20, Q21 and Q25 (Part I pp. 117–118). Q20 asks about a square minus a sine plus a constant at the input pi. Q21 asks about the sum, the difference and the product of sine and cosine. Q25 is a difference of sine and cosine away from zero, with minus one assigned at zero. Verified: Q20 is continuous there, being a sum of continuous functions, and its value is the square of pi plus five, since the sine of pi is zero. All three parts of Q21 are continuous everywhere, by the first three parts of Theorem 1. Q25 is continuous everywhere: away from zero by Theorem 1, and at zero because the limit there is minus one, which is precisely the value assigned.
• Exercise 5.1 Q22 (Part I p. 117). Cosine, cosecant, secant and cotangent. Verified: the cosine is unbroken across the whole line. The cosecant is the reciprocal of the sine, continuous except at the whole multiples of pi. The secant is the reciprocal of the cosine, continuous except at the odd multiples of half pi. The cotangent is a quotient with the sine below, continuous except at the whole multiples of pi. All four follow from Theorem 1 and the second Remark, and the excluded sets are exactly the zeros of the denominators.
• Exercise 5.1 Q31, Q32 and Q33 (Part I p. 118). A cosine of a square, a modulus of a cosine, and a sine of a modulus. Verified: each stacks two rules already known unbroken everywhere, so Theorem 2 settles all three in one line apiece. The pair Q32 and Q33 is worth pausing on: the same two functions in the two possible orders, both continuous, and the two graphs are quite different.
• Exercise 5.1 Q24 (Part I p. 117). A square times the sine of a reciprocal away from zero, and zero at zero. Verified: away from zero, the reciprocal and the sine and the square are all continuous and Theorem 2 with Theorem 1 settles it. At zero the two theorems say nothing, because the inner reciprocal is not defined there, let alone continuous. The verdict is still continuity, but it needs a bounding argument — the whole expression is trapped between the square and its negative — and the chapter states no such theorem. This is the closing item of the explanation and it should be presented as the boundary of the method, not as an exercise like the others.

Figures to have open

• A four-row table of the parts of Theorem 1, with columns for the combination and its proviso. The content is the chapter's; the layout is added here. Use the repo's DataTable component.
• An annotated redraw of the sum proof (Part I p. 113) with the chapter's own four margin justifications kept as labels. Nothing here is a figure in the book; this is an added device.
• Two number lines for section 9, one carrying the whole multiples of pi and one the odd multiples of half pi, drawn to the same scale so that the two excluded sets can be seen to interleave.
• A two-box series diagram for sections 10 and 11, with the two continuity demands pinned to the input of the first box and the input of the second.
• Graphs of a modulus of a cosine and a sine of a modulus for section 12, drawn on the same axes range so the difference in shape is visible. The chapter draws neither; there is no figure anywhere in §5.2.1 or in Part I pp. 113–125.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 5, §5.2.1 Algebra of continuous functions, opening paragraph and Theorem 1 with its proof, Part I p. 113
• Remarks (i) and (ii) and Example 16, Part I p. 114; Example 17 and the fact it declares, Part I p. 114
• Remark on the cosine, Example 18, Theorem 2 and Examples 19 and 20, Part I p. 115
• Exercise 5.1, questions 20 to 25, Part I pp. 117–118
• Exercise 5.1, questions 31 to 33, Part I p. 118; Summary, the algebra bullet, Part I p. 146