Trapping a function between two others to settle the trigonometric cases

What to assume they know

• Limits pass through sums, products and quotients — Theorem 1, and in particular the product and quotient parts
• Substitution works until it gives nothing over nothing, and then cancellation does — the 0/0 diagnosis and why cancellation is legal
• Radian measure, and the area of a circular sector as a fraction of the whole circle
• The area of a triangle as half base times height
• sin, cos and tan of an angle read off a unit circle, and the identity relating sin², cos² and 1
• The identity for 1 − cos x in terms of a half-angle sine, from Shifts by a quarter, a half and a whole turn all fall out of the same two results
• The identity for sin A − sin B, used again in Sine and tangent go back to the definition, because no rule so far reaches them

What they should be able to do

• State Theorem 3 and say exactly what it claims about two functions that are ordered on their common domain
• State the Sandwich Theorem, naming its three functions, the ordering they must satisfy, and the two limits that must agree
• Build the three-region comparison in the unit circle and identify which triangle, which sector and which triangle bound each other
• Convert that area comparison into the chain sin x < x < tan x for positive x below π/2
• Divide through, take reciprocals, and obtain the printed inequality, keeping the direction of each inequality correct through both operations
• Explain why the inequality is quoted with |x| and how the negative side is covered
• Prove the limit of sin x over x at 0 by the Sandwich Theorem
• Reduce the limit of (1 − cos x)/x at 0 to the first one using the half-angle identity, and justify the change of variable it needs
• Evaluate composite trigonometric limits by rewriting them into copies of the two standard limits

Where it usually goes wrong

• "sin x / x at 0 is 1 because sin x is about x for small x." That is the conclusion restated, not an argument. The chapter's inequality is what makes "about" precise, and the whole point of §12.4 is that the approximation needs proving.
• "The Sandwich Theorem needs the two outer functions to be equal." They must share a limit at the one point. Everywhere else they can be as far apart as they like. On the interval the chapter's inequality is asserted over — the punctured range out to a quarter turn either side of 0 — the cosine and the constant 1 meet nowhere at all. Say the interval out loud: over the whole real line they meet at every whole number of full turns, so the claim is false without the restriction.
• "Taking reciprocals leaves an inequality alone." It reverses it, and only when everything in sight is positive. Both facts are used at once on p. 235, and both are places where students silently go wrong.
• "The inequality holds for all x." It is asserted on a punctured interval around 0 of radius π/2, with 0 itself excluded because sin x / x has no value there. Outside that interval it is not claimed.
• "The x in the sector area is a length." It is the angle in radians, and the sector formula is only that simple because of radian measure. In degrees the whole argument collapses and the limit is not 1.
• "The second limit needs its own inequality." It does not — it is the first limit plus one identity. Reducing rather than reproving is the move worth teaching.
• "sin x / x behaves differently for negative x." It does not, and the evenness argument on p. 235 is the reason. It is also the reason the printed condition uses a modulus.
• "Because 0 < |x| < π/2 is where the inequality holds, the limit statement is restricted too." A limit at 0 only ever consults values near 0. The interval is generous for that purpose.

Questions to check understanding

• State the Sandwich Theorem and use it on a supplied inequality
• Prove the inequality relating cos x, sin x / x and 1 from the unit-circle areas
• Evaluate sin ax / bx, sin ax / sin bx and tan kx / x type limits at 0
• Evaluate a limit at a point other than 0 by a substitution that moves it to 0 — Exercise 12.1 q15 and q22 are both of this kind
• Rewrite a limit involving cosec or cot into sines and cosines and evaluate it
• Two-mark: explain why the standard trigonometric limits require radian measure
• Derive (1 − cos x)/x² at 0 as an extension, using the same half-angle identity

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 3 (§12.4, p. 234). Two real valued functions on one domain, with the first never exceeding the second anywhere on it. If both limits exist at a point, then the first limit does not exceed the second. The comparison signs are inclusive in both the hypothesis and the conclusion — read off the p. 234 page image, not the text layer. Illustrated by Fig 12.8.
• Fig 12.8 (p. 234), read off the page image. Two bell-shaped curves on one pair of axes, the upper labelled y = g(x) and the lower y = f(x), with a vertical dashed line at the marked abscissa a. The picture makes the point that the ordering holds everywhere, not only at a.
• Theorem 4, the Sandwich Theorem (§12.4, p. 234). Three real valued functions on a common domain, ordered so that the middle one lies between the outer two everywhere. If the outer two share a common limit at a point, the middle one has that same limit. Inclusive signs again, read off the page image. Illustrated by Fig 12.9: three curves, labelled from the top y = h(x), y = g(x), y = f(x), converging to a single height above the marked a.
• **The inequality, printed as (*) (p. 234): cos x < sin x / x < 1, asserted for 0 < |x| < π/2. The modulus bars are printed and do not survive extraction — p234.txt returns the condition without them. Read off the page image.** Both inner comparisons are strict.
• Fig 12.10 (p. 235), read off the page image. A unit circle centred at O, with A on the circle to the right on the horizontal radius and C on the circle above it, the angle AOC equal to x radians with x between 0 and π/2. D is the foot of the perpendicular from C to OA. B lies on the line through O and C extended, positioned so that BA is perpendicular to OA — so B sits outside the circle, directly above A. The chord AC is drawn.

The shading, checked on the printed page, is not the sector. What carries the tint is the whole disc, together with the piece of triangle OAB that spills outside the circle — so the tinted region is the disc and that spill taken together. The sector the argument turns on is not picked out by shading at all, which is worth knowing precisely because the explanation will want to pick it out.

• The three areas (p. 235). Triangle OAC, then sector OAC, then triangle OAB, in increasing order. Written out with the printed expressions: half of OA times CD; the fraction x over 2π of the whole disc area π(OA)²; and half of OA times AB. Verified: cancelling the common factor of OA from all three, and using CD = OA·sin x and AB = OA·tan x from the two right triangles, the chain becomes sin x < x < tan x. The middle term simplifies because x/(2π) × π(OA)² = ½·x·(OA)², and every term now carries ½·OA·(OA), which is positive.
• The two manipulations (p. 235). Dividing the chain by sin x, which is positive on this range, preserves the order and gives 1 < x/sin x < 1/cos x. Taking reciprocals of a chain of positive quantities reverses it and gives the printed (). Verified:* both steps are legitimate only because every quantity involved is positive on 0 < x < π/2, which is why the chapter reduces to that range at the start of the proof.
• The reduction to positive x (p. 235). The proof opens by recording how each of the two functions responds to a change of sign in its input — sine reverses, cosine does not — and concludes on that basis that arguing for positive x is enough. Verified: those two facts make sin x / x unchanged under x → −x, since both the numerator and the denominator change sign, and cos x is unchanged as well. So the whole of (*) transfers to the negative side unaltered. This is why the statement carries |x| rather than x, and an explanation that drops the bars has quietly proved half the result.
• Theorem 5 (i) (p. 235). The limit of sin x over x at 0 is 1. Proof as printed: (*) traps sin x / x between cos x and the constant 1; the limit of cos x at 0 is 1; the Sandwich Theorem finishes it.
• Theorem 5 (ii) (pp. 235–236). The limit of (1 − cos x)/x at 0 is 0. Proof as printed: use 1 − cos x = 2sin²(x/2), so the quotient becomes [sin(x/2)/(x/2)]·sin(x/2); the first factor tends to 1 by part (i) and the second to 0. Verified: 2sin²(x/2)/x = [sin(x/2)/(x/2)]·sin(x/2) exactly, and 1 × 0 = 0. The chapter flags that it has used the equivalence of x → 0 and (x/2) → 0, justified by the substitution y = x/2.
• Example 4 (pp. 236–237). (i) sin 4x / sin 2x at x → 0. Verified: rewrite as (sin 4x / 4x) × (2x / sin 2x) × 2; each bracket tends to 1 as its own argument tends to 0, giving 2. (ii) tan x / x at x → 0. Verified: tan x / x = (sin x / x)(1/cos x), giving 1 × 1 = 1.
• Exercise 12.1 items this topic owns (pp. 237–238): 13 to 22. Item 13 is sin ax / bx; item 14 is sin ax / sin bx with a and b non-zero; item 15 is sin(π − x)/(π(π − x)) at x → π; item 16 is cos x/(π − x) at x → 0; item 17 is (cos 2x − 1)/(cos x − 1); item 18 is (ax + x cos x)/(b sin x); item 19 is x sec x; item 20 is (sin ax + bx)/(ax + sin bx) with a, b and a + b non-zero; item 21 is cosec x − cot x; item 22 is tan 2x/(x − π/2) at x → π/2.

Figures to have open

• Fig 12.10 redrawn (p. 235) at large scale: unit circle, the radius OA, the point C on the circle, the perpendicular CD onto OA, the point B outside the circle with BA perpendicular to OA, the chord AC, and the angle labelled x radians. The printed shading covers the disc and the spill of triangle OAB beyond it, not the sector — a redraw may highlight the sector instead, since that is what the argument needs, but it should do so knowingly rather than by copying a shading it has misread. This is the figure the whole topic rests on and it must be drawn accurately, with B clearly outside the circle — the drawing is what makes the third area the largest. Read off the p. 235 page image.
• Figs 12.8 and 12.9 redrawn (p. 234) as schematics.
• A three-panel area comparison: the inner triangle, then the sector, then the outer triangle, drawn to the same scale so the ordering is visible rather than asserted.
• A graph panel carrying y = cos x, y = 1 and y = sin x / x together on −π/2 < x < π/2, with the point at 0 removed from the third curve. Standard schematic; the chapter prints no such graph.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 12 "Limits and Derivatives", §12.4 Limits of Trigonometric Functions, printed pp. 234–237.
• Theorem 3 and Theorem 4 with Figs 12.8 and 12.9 (p. 234); the inequality labelled (*) (p. 234); its geometric proof with Fig 12.10 (p. 235); Theorem 5 and its proof (pp. 235–236); Example 4 (pp. 236–237); the general recipe closing the section (p. 237).
• Exercise 12.1 items 13–22, printed pp. 237–238.
• Chapter Summary, p. 255, which lists both standard limits.
• Deliberate cross-reference outside this chapter: the half-angle identity for 1 − cos x and the sum-and-difference identities used in Example 4 belong to Chapter 3 and are used here without re-derivation.