Lesson 17: Functions and Limits

Class XI mathematics begins where Class X ends: with the idea of a function as a machine that assigns to each input exactly one output, and the idea of a limit — the value a function approaches as its input approaches a point, whether or not it ever gets there. These two ideas are the entire foundation of calculus. This lesson builds them carefully, in the NCERT spirit.

Functions: notation and evaluation

Writing f(x) = x² − 3x + 1 means the machine "square, triple-subtract, add 1". Evaluation is substitution: f(2) = 4 − 6 + 1 = −1; f(−1) = 1 + 3 + 1 = 5; f(a + h) = (a + h)² − 3(a + h) + 1 — and that last, unbracketed-carefully form is the one calculus is made of: the difference quotient [f(a + h) − f(a)]/h measures average rate of change, and Lesson 18 is the story of what happens as h shrinks.

Two disciplines matter from day one. First, the domain: f(x) = 1/(x − 2) is defined for all x ≠ 2; f(x) = √(x − 3) needs x ≥ 3. Second, composition order: f(g(x)) means g first, then f — and f∘g ≠ g∘f in general.

Limits: approaching, not arriving

lim (x→a) f(x) = L means: as x gets arbitrarily close to a (from either side), f(x) gets arbitrarily close to L. The definition deliberately ignores f(a) itself — f(a) may be different, or may not exist at all.

The fundamental example. Consider f(x) = (x² − 4)/(x − 2), undefined at x = 2. Values near 2: at x = 1.9, f = 3.9; at 1.99, 3.99; at 2.01, 4.01. The function *approaches* 4. And algebra explains why: (x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = x + 2 for all x ≠ 2, so the limit as x → 2 is 4. Factor-and-cancel is the first technique of limits: the cancelled factor is exactly the one killing the function at the point.

Worked example 1 (basic). lim (x→3) (x² − 9)/(x − 3) = lim (x + 3) = 6, by the same factorisation.

Worked example 2 (intermediate — rationalising). lim (x→0) (√(x + 1) − 1)/x. Direct substitution gives 0/0. Multiply top and bottom by the conjugate √(x + 1) + 1: numerator becomes (x + 1) − 1 = x; the fraction is x/[x(√(x + 1) + 1)] = 1/(√(x + 1) + 1) for x ≠ 0, and the limit is 1/(1 + 1) = 1/2.

Worked example 3 (exemplar style). lim (x→1) (x³ − 1)/(x − 1). Factor: x³ − 1 = (x − 1)(x² + x + 1), so the limit is 1 + 1 + 1 = 3. This generalises: lim (x→a) (xⁿ − aⁿ)/(x − a) = n·aⁿ⁻¹ — a standard result NCERT Class XI expects you to *use*; notice it is secretly a derivative (Lesson 18) in disguise.

The trigonometric limit and behaviour at infinity

Two more workhorses:

• lim (x→0) (sin x)/x = 1 — with x in radians. (In degrees the value is π/180; calculus uses radians precisely so this limit is the clean number 1.)
• Limits at infinity: for (3x² + 2x)/(x² − 5) as x → ∞, divide every term by x²: (3 + 2/x)/(1 − 5/x²) → 3. The highest powers rule the far behaviour; polynomials of equal degree settle to the ratio of leading coefficients.

Left and right limits. A two-sided limit exists only if both one-sided limits agree. For f(x) = |x|/x near 0: from the right f = 1, from the left f = −1 — no two-sided limit. When a problem says "discuss the limit", this check is what it wants.

Competency application: average speed becomes instantaneous

A ball's displacement is s(t) = t² metres. Its *average* velocity between t = 3 and t = 3 + h is [(3 + h)² − 9]/h = (6h + h²)/h = 6 + h. As h → 0, this tends to 6 m/s — the instantaneous velocity at t = 3. Competency sets dress this as a drone's altitude or a car's odometer, but the skeleton is identical: form the difference quotient, cancel h, then let h → 0. You have just performed the central computation of the next lesson.

Common mistakes

• Substituting and declaring 0/0 = "no limit". 0/0 is *indeterminate* — a signal to factor, rationalise, or use a standard limit, never a conclusion.
• Cancelling (x − a) and then claiming the functions are identical. They agree except at x = a — which is precisely why limits can differ from values, and why the technique works.
• Using sin x/x = 1 with degrees. The identity needs radians.
• At infinity, "cancelling" the biggest terms casually. Divide numerator and denominator by the highest power systematically; guessing by "the x² terms dominate, so 3/1" happens to work here but the method is what generalises.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. lim (x→2) (3x² − x − 10)/(x − 2) equals: (a) 0 (b) 11 (c) 7 (d) does not exist. Answer: (b) 11. 3x² − x − 10 = (x − 2)(3x + 5); cancelling gives 3x + 5 → 11 at x = 2.

Q2. lim (x→0) (sin 5x)/x equals: (a) 0 (b) 1 (c) 5 (d) 1/5. Answer: (c) 5. Write (sin 5x)/x = 5 × (sin 5x)/(5x) and use the standard limit with 5x → 0.

Section B — Short answer (3 marks) Q3. Evaluate lim (x→∞) (3x² + 2x)/(x² − 5). Marking scheme: 1 mark for dividing numerator and denominator by x²; 1 mark for obtaining (3 + 2/x)/(1 − 5/x²); 1 mark for the limit 3, noting 2/x and 5/x² → 0.

Section C — Case-based long answer (5 marks) Q4. A drone's height t seconds after launch is h(t) = 2t² + 3t metres. (i) Find its average velocity between t = 1 and t = 1 + h. (2 marks) (ii) Hence find its instantaneous velocity at t = 1. (1 mark) (iii) A second drone follows H(t) = t² + 8t. At t = 1, which drone is climbing faster, and by how much? (2 marks) Marking scheme: (i) Difference quotient: [2(1 + h)² + 3(1 + h) − 5]/h = [2 + 4h + 2h² + 3 + 3h − 5]/h = (7h + 2h²)/h = 7 + 2h (1 mark for expansion, 1 for cancelling h). (ii) As h → 0: 7 m/s (1 mark). (iii) For H: quotient at t = 1 is [(1 + h)² + 8(1 + h) − 9]/h = (10h + h²)/h = 10 + h → 10 m/s (1 mark); the second drone is faster by 3 m/s (1 mark).

Going deeper: the standard limit lim (x→a) (xⁿ − aⁿ)/(x − a) = n·aⁿ⁻¹

Worked example 3 met this for n = 3, a = 1. Apply it: lim (x→2) (x³ − 8)/(x − 2) = 3 × 2² = 12. Direct verification: x³ − 8 = (x − 2)(x² + 2x + 4), and the second factor at x = 2 is 4 + 4 + 4 = 12 ✓. Two more from the same family: lim (x→3) (x² − 9)/(x − 3) = 2 × 3 = 6 (Lesson 17's first example in hindsight), and lim (x→1) (x⁵ − 1)/(x − 1) = 5 × 1⁴ = 5. Keep the pattern in view: this limit *is* the derivative of xⁿ at a, which is why Lesson 18's power rule has the same shape — limits and derivatives are one subject, not two.

Going deeper: domain and range, worked properly

For f(x) = √(9 − x²): the radicand needs 9 − x² ≥ 0, so the domain is [−3, 3]. The values of √(9 − x²) run from 0 (at x = ±3) up to 3 (at x = 0), so the range is [0, 3]. (The graph is the upper semicircle of radius 3 — domain and range are its shadow on each axis.) Contrast f(x) = 1/(x − 2): domain all reals except 2; range all reals except 0 — a value is excluded from the range when no input can produce it, here because 1/(anything) is never 0. Domain questions ask "what may go in"; range questions ask "what can come out" — answering one with the other's reasoning is the standard slip.

Going deeper: continuity — when the limit meets the value

A function is continuous at a when lim (x→a) f(x) = f(a): the journey and the destination agree. The Lesson 9-style troublemaker f(x) = (x² − 4)/(x − 2) for x ≠ 2 has limit 4 at x = 2; defining f(2) = 4 makes it continuous (the hole is "removable"), while defining f(2) = 5 leaves a jump between limit and value. Piecewise definitions at the seam are the standard test: check the left limit, the right limit, and the value — all three must coincide. Continuity is not examined in depth until later classes, but the three-way check is the entire concept, and it is already within reach.

Final drills

Drill 1 — two standard limits worth owning. (a) lim (x→0) (1 − cos x)/x² = 1/2: write 1 − cos x = 2 sin²(x/2), so the expression is 2 sin²(x/2)/x² = (1/2) × [sin(x/2)/(x/2)]² → (1/2) × 1. (b) lim (x→4) (√x − 2)/(x − 4): rationalise by (√x + 2): the numerator becomes x − 4, cancelling to 1/(√x + 2) → 1/4. Both are the lesson's two moves — standard limits and conjugates — in their most reusable forms.