Two standard limits, the first squeezed out of an inequality and the second reduced to it

What to assume they know

• The exponential and the logarithm, their domains, ranges and graphs — the two functions, their domains, ranges and graphs
• Trapping a function between two others to settle the trigonometric cases — the Sandwich Theorem and the two trigonometric standard limits
• Substitution works until it gives nothing over nothing, and then cancellation does — the 0/0 diagnosis, and evaluating a limit by changing the variable
• Index laws, in particular that e raised to a sum splits into a product
• The identity relating 1 − cos x to a half-angle sine
• Reading a chain of inequalities and knowing which comparisons are inclusive

What they should be able to do

• State the inequality §12.6 uses, including the interval on which it is asserted and the point excluded from it
• State the first standard limit and prove it by the Sandwich Theorem from that inequality
• Show that each of the two bounding expressions tends to 1 at 0, and say which fact about limits each of those two computations uses
• State the second standard limit and derive it from the first by a substitution, naming the substitution and why the new variable also tends to 0
• Evaluate a limit in which the exponent is a constant multiple of x, by matching the argument to the denominator
• Evaluate a limit of a difference by splitting it into two standard limits
• Evaluate a limit at a point other than 0 by shifting the variable
• Combine an exponential limit with a trigonometric one in a single computation

Where it usually goes wrong

• "The inequality is proved on the page." It is not. It is stated and used. This is a real difference from §12.4, where the corresponding inequality gets a full geometric proof, and the contrast is worth drawing.
• "Because the inequality holds only between −1 and 1, the limit statement is restricted." A limit at 0 only ever consults values close to 0; an interval of radius 1 is more than enough.
• "The comparisons in the inequality are strict, like the trigonometric one." They are inclusive here, and strict in §12.4. Both were read off the page images. If an explanation draws them wrongly, the two arguments stop looking like the same technique applied twice, which is the whole point of putting them together.
• "The logarithm limit needs its own inequality." It needs a substitution. One inequality supports both theorems, and noticing that is the section's economy.
• "(e^{3x} − 1)/x tends to 1, because that is the standard limit." The standard limit needs the exponent's argument and the denominator to be the same thing. When they differ by a constant factor, that factor comes out. Item 1 of the exercise and Example 5 are both testing exactly this.
• "e^{2+x} − e² cannot be handled, because the standard limit starts from 1." Splitting the exponential of a sum into a product pulls out e² and leaves precisely the standard form. Two of the eight exercise items are that move.
• "The section teaches derivatives of the exponential and the logarithm." It does not. It treats limits only. No derivative of either function appears there.

Questions to check understanding

• Evaluate a limit of the form (e^{kx} − 1)/x at 0
• Evaluate a limit of the form (eˣ − eᵃ)/(x − a) at a
• Evaluate a limit of log(1 + kx)/x at 0
• Evaluate a limit at a point other than 0 by shifting the variable
• Evaluate a limit combining an exponential quotient with a trigonometric one
• State the Sandwich Theorem and identify the two bounding functions in a supplied argument
• Two-mark: explain why the second standard limit does not require a second inequality

Examples worth working on the board

Values marked verified are worked out here from what the section states; no answer key was consulted, and the section's own answers are addressed in the Notes below.

• The inequality (Supplementary Material, p. 360). The exponential difference quotient is trapped between two expressions. The lower bound is the reciprocal of 1 plus the modulus of x. The upper bound is 1 plus the modulus of x multiplied by (e − 2). Both comparisons are inclusive, and the assertion is made for every x in the closed interval from −1 to 1 with 0 removed. Read off the p. 360 page image. The extracted text loses the modulus bars in the lower bound's denominator, turning it into a different and false statement; take this from the image only.
• What is not proved (section 3). The section introduces the inequality with no derivation and no reference. It is used twice on the page and never established. Say so; an explanation that presents it as proved is misrepresenting the source.
• The two ends (pp. 360–361). Verified: the lower bound tends to 1 because the modulus of x tends to 0, so the denominator tends to 1 and the quotient rule of Theorem 1 applies with a non-zero denominator limit. The upper bound tends to 1 because (e − 2) is a fixed number and the modulus of x tends to 0, so the product tends to 0 and the sum to 1. Two different parts of Theorem 1 are doing the work at the two ends; name which.
• Theorem 6 of the Supplementary Material (p. 360). The limit of the exponential difference quotient at 0 is 1, proved by the Sandwich Theorem from the inequality above. This theorem is printed as Theorem 6, which is also the number carried by the power rule in §12.5 at p. 245. Both were confirmed on the page images. Cite by page.
• Theorem 7 of the Supplementary Material (p. 361). The limit of the logarithm of (1 + x), divided by x, at 0 is 1. The printed proof sets that quotient equal to a new variable y, rearranges to express 1 + x as e raised to the product of x and y, and thereby produces the first theorem's quotient with that product as its argument — from which y is forced to tend to 1. Also numbered 7, colliding with the polynomial-derivative theorem at §12.5.2, p. 246. Verified by an independent route the explanation may prefer: put u = log(1 + x), so that 1 + x = eᵘ and x = eᵘ − 1, and u tends to 0 as x does; then the quotient is u divided by (eᵘ − 1), which is the reciprocal of Theorem 6's quotient and so tends to 1. This is shorter than the printed argument and reaches the same place. If the explanation uses it, say it is not the book's.
• Example 5 (p. 361). The limit at 0 of (e^{3x} − 1)/x. Verified: multiply and divide by 3 so the denominator matches the exponent's argument; as x tends to 0 so does 3x, and the bracket becomes Theorem 6's quotient, giving 3 × 1 = 3.
• Example 6 (p. 362). The limit at 0 of (eˣ − sin x − 1)/x. Verified: split into (eˣ − 1)/x minus sin x/x. The first tends to 1 by Theorem 6, the second to 1 by §12.4's Theorem 5 (i), so the answer is 0. This example is the one place the section reaches back into the chapter proper.
• Example 7 (p. 362). The limit at 1 of (log x)/(x − 1). Verified: put x = 1 + h, so that h tends to 0 as x tends to 1 and the expression becomes log(1 + h)/h, which is Theorem 7's quotient. The value is 1.
• The exercise (p. 362), eight items. Printed under a heading reading Exercise 13.2 — a chapter-13 leftover; Chapter 12's own last exercise is Exercise 12.2 at pp. 248–249, so anyone told to look for "Exercise 12.3" will find nothing. The items:
  • (e^{4x} − 1)/x at 0. Verified: 4.
  • (e^{2+x} − e²)/x at 0. Verified: factor e² out of the numerator, leaving e² times Theorem 6's quotient, so e².
  • (eˣ − e⁵)/(x − 5) at 5. Verified: factor e⁵ out and substitute h = x − 5, giving e⁵.
  • (e^{sin x} − 1)/x at 0. Verified: write it as [(e^{sin x} − 1)/sin x] × (sin x / x); the first bracket is Theorem 6 with argument sin x, which tends to 0, and the second is §12.4's limit, so 1.
  • (eˣ − e³)/(x − 3) at 3. Verified: e³, by the same route as item 3.
  • x(eˣ − 1)/(1 − cos x) at 0. Verified: write it as [(eˣ − 1)/x] × x²/(1 − cos x). Using 1 − cos x = 2sin²(x/2), the second factor is x² divided by 2sin²(x/2), which is 2 divided by the square of [sin(x/2)/(x/2)] — and that tends to 2. So the answer is 1 × 2 = 2. This item needs the half-angle identity and the trigonometric limit as well as the exponential one, and it is the hardest thing in the section.
  • log(1 + 2x)/x at 0. Verified: multiply and divide by 2, giving 2 × Theorem 7's quotient with argument 2x, so 2.
  • log(1 + x³)/sin³x at 0. Verified: write it as [log(1 + x³)/x³] × [x/sin x]³. The first tends to 1 by Theorem 7 with argument x³, the second cube tends to 1, so the answer is 1.

Figures to have open

• A three-curve squeeze plot on the interval from −1 to 1: the two bounds and the quotient between them, with the point at 0 removed. The section prints no such graph; this is added here, and section 5 needs it.
• A number line showing the closed interval from −1 to 1 with 0 punched out, and the inclusive endpoints drawn as filled markers. The distinction between this and §12.4's strictly bounded punctured interval is the point.
• A side-by-side of the two squeeze arguments — §12.4's trigonometric one and §12.6's exponential one — with the proved inequality on one side and the asserted one on the other.
• A grouping board for the eight exercise items by technique: constant factor in the exponent; exponential of a sum; shift of the point; composition with another function.
• No printed figure from the source is needed for this topic. Fig. 13.11 and Fig. 13.12 belong to The exponential and the logarithm, their domains, ranges and graphs.

Where this sits in the book

• NCERT Class XI Mathematics, Supplementary Material, §12.6 "Limits Involving Exponential and Logarithmic Functions", printed pp. 359–362, under a heading giving the chapter number on p. 359. The inequality and the first theorem are on p. 360, the first theorem's proof completes on p. 361 with the second theorem beneath it, and Examples 5 to 7 with the exercise occupy pp. 361–362.
• The exercise is printed as Exercise 13.2 on p. 362.
• Backward references into the chapter proper: the Sandwich Theorem is Theorem 4, §12.4, p. 234; the trigonometric limits used by Examples 6 and by exercise items 4, 6 and 8 are Theorem 5, §12.4, p. 235; the half-angle identity needed by item 6 is used at §12.4, p. 236.
• The exponential and the logarithm, their domains, ranges and graphs covers the two functions themselves.