Limits and Derivatives

2.1 Core concepts

• Calculus studies how the value of a function changes as the points in its domain change. The core ideas covered are: the intuitive notion of a derivative, the (naive) limit, the algebra of limits, the formal derivative, and the standard derivative formulae (NCERT §12.1, p. 217).
• The free-fall law s = 4.9t² with successive average velocities computed over shrinking intervals ending at and starting from t = 2 yields a common limiting value, called the instantaneous velocity at t = 2; this is the prototype of a derivative (NCERT §12.2, Tables 12.1–12.3, p. 218–219).
• Geometrically, the sequence of chords C_iB_i / AC_i approaches the slope of the tangent at A as the time intervals shrink to zero, so the instantaneous velocity equals the slope of the tangent to s = 4.9t² at t = 2 (NCERT §12.2, Fig 12.1, p. 219–220).
• For a function f, lim_{x→a} f(x) = l means f(x) approaches l as x approaches a; x may approach a from the left or from the right, giving left-hand limit lim_{x→a⁻} f(x) and right-hand limit lim_{x→a⁺} f(x). The limit exists at x = a if and only if these two one-sided limits coincide (NCERT §12.3, p. 220–221).
• The example f(x) = 1 for x ≤ 0 and f(x) = 2 for x > 0 shows that the LHL (= 1) and RHL (= 2) at 0 can differ, in which case lim_{x→0} f(x) does not exist even though f(0) is defined (NCERT §12.3, Fig 12.3, p. 221).
• In general the value of the function at a point and its limit at that point need not be equal: e.g. f(x) = x + 2 for x ≠ 1, f(1) = 0 gives lim_{x→1} f(x) = 3 ≠ f(1) (NCERT §12.3 Illustration 10, Fig 12.7, p. 227).
• Algebra of limits (Theorem 1): If lim f(x) and lim g(x) both exist at a, then lim[f ± g] = lim f ± lim g; lim[f·g] = lim f · lim g; lim[f/g] = lim f / lim g provided lim g ≠ 0; and lim(λ·f) = λ·lim f for any constant λ (NCERT §12.3.1, p. 228).
• Polynomials: Using lim_{x→a} x = a and induction, lim_{x→a} xⁿ = aⁿ; therefore for any polynomial f, lim_{x→a} f(x) = f(a) (NCERT §12.3.2, p. 228–229).
• Rational functions: For f(x) = g(x)/h(x) with h(a) ≠ 0, lim_{x→a} f(x) = g(a)/h(a). If h(a) = 0 and g(a) ≠ 0, the limit does not exist. If both vanish, cancel common (x − a) factors and re-evaluate; the limit is 0 if the order of vanishing of g exceeds that of h, undefined if it is less (NCERT §12.3.2, p. 229–230).
• Standard algebraic limit (Theorem 2): For any positive integer n, lim_{x→a} (xⁿ − aⁿ)/(x − a) = n·a^(n−1). The result extends to any rational exponent n with a > 0 (NCERT §12.3.2 Remark, p. 232–233).
• Sandwich Theorem (Theorem 4): If f(x) ≤ g(x) ≤ h(x) on a common domain and lim_{x→a} f = lim_{x→a} h = l, then lim_{x→a} g = l (NCERT §12.4, p. 234).
• The geometric inequality cos x < sin x / x < 1 for 0 < x < π/2 is established from the circle figure (area of triangle OAC < area of sector OAC < area of triangle OAB) (NCERT §12.4, Fig 12.10, p. 235).
• Standard trigonometric limits (Theorem 5): lim_{x→0} sin x / x = 1 and lim_{x→0} (1 − cos x)/x = 0. The first follows from the sandwich theorem; the second uses the identity 1 − cos x = 2 sin²(x/2) (NCERT §12.4, p. 235–236).
• Derivative at a point (Definition 1): f′(a) = lim_{h→0} [f(a+h) − f(a)]/h, provided the limit exists; f′(a) measures the rate of change of f at a (NCERT §12.5, p. 240).
• Derivative as a function (Definition 2 — first principle): f′(x) = lim_{h→0} [f(x+h) − f(x)]/h wherever the limit exists; also denoted d/dx (f(x)), dy/dx, or D(f(x)) (NCERT §12.5, p. 242).
• Geometric meaning: f′(a) equals tan ψ, the slope of the tangent to y = f(x) at (a, f(a)); the chord PQ tends to this tangent as h → 0 (NCERT §12.5, Fig 12.11, p. 241–242).
• Algebra of derivatives (Theorem 5 of §12.5.1): (u ± v)′ = u′ ± v′; product (Leibnitz) rule (uv)′ = u′v + uv′; quotient rule (u/v)′ = (u′v − uv′)/v², valid where v ≠ 0 (NCERT §12.5.1, p. 244).
• Power rule (Theorem 6): d/dx (xⁿ) = n x^(n−1) for any positive integer n (proved via the binomial expansion of (x + h)ⁿ); the remark extends it to all real n (NCERT §12.5.1, p. 245–246).
• Polynomial derivative (Theorem 7): d/dx (aₙxⁿ + … + a₁x + a₀) = n aₙ x^(n−1) + (n−1) a_{n−1} x^(n−2) + … + a₁ (NCERT §12.5.2, p. 246).
• Standard trig derivatives: d/dx (sin x) = cos x (Example 16, p. 247); d/dx (cos x) = − sin x (Summary, p. 255); d/dx (tan x) = sec² x (Example 17, p. 247–248); d/dx (cot x) = − cosec² x (Example 21, p. 251–252).

2.2 Definitions to memorise

Standard limits to memorise (Summary, p. 254–255):

Standard derivatives (Summary, p. 255): d/dx (xⁿ) = n x^(n−1); d/dx (sin x) = cos x; d/dx (cos x) = − sin x.

2.3 Diagrams / processes to remember

• Fig 12.1 (p. 219): Plot of s = 4.9t² with shrinking time intervals h₁, h₂, …; the chord slopes C_iB_i/AC_i tend to the slope of the tangent at A — the geometric origin of the derivative.
• Fig 12.2 (p. 220): Graph of h(x) = (x² − 4)/(x − 2), an undefined point at x = 2 with limit 4 — shows that limit can exist where the function value does not.
• Fig 12.3 (p. 221): Step-like function with f(x) = 1 for x ≤ 0 and 2 for x > 0; visual demonstration that LHL ≠ RHL ⇒ limit does not exist.
• Fig 12.6 (p. 227): Piecewise function with jump at 0 — LHL = −2, RHL = +2, so limit does not exist though f(0) = 0.
• Fig 12.7 (p. 227): f(x) = x + 2 for x ≠ 1, f(1) = 0 — limit at 1 is 3, value is 0; limit ≠ value.
• Fig 12.10 (p. 235): Unit circle with sector OAC; the area inequality (1/2)OA·CD < (1/2)π(OA)²(x/π) < (1/2)OA·AB delivers cos x < sin x / x < 1.
• Fig 12.11 (p. 241): Secant PQ through (a, f(a)) and (a+h, f(a+h)); as h → 0, secant tends to tangent at P, so slope (QR/PR) → f′(a).

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 2(ii), p. 230). Evaluate lim_{x→2} (x³ − 4x² + 4x)/(x² − 4).

Step 1 — factor: numerator = x(x − 2)²; denominator = (x − 2)(x + 2). Step 2 — cancel: limit = lim_{x→2} x(x − 2)/(x + 2). Step 3 — substitute: 2·0/4 = 0.

Example B (NCERT Example 3(i), p. 233). Evaluate lim_{x→1} (x¹⁵ − 1)/(x¹⁰ − 1).

Step 1 — rewrite as ratio of standard limits: [(x¹⁵ − 1)/(x − 1)] ÷ [(x¹⁰ − 1)/(x − 1)]. Step 2 — apply Theorem 2: 15·1¹⁴ / 10·1⁹ = 15/10. Step 3 — simplify: = 3/2.

Example C (NCERT Example 4(i), p. 236). Evaluate lim_{x→0} sin 4x / sin 2x.

Step 1 — rewrite: [(sin 4x)/(4x)] · [(2x)/(sin 2x)] · 2. Step 2 — limits: each ratio → 1. Step 3 — product: 1·1·2 = 2.

Example D (NCERT Example 5, p. 240). First-principle derivative of f(x) = 3x at x = 2.

Step 1 — set up: f′(2) = lim_{h→0} [3(2+h) − 3·2]/h. Step 2 — simplify numerator: 3h. Step 3 — limit: 3h/h = 3 ⇒ f′(2) = 3.

Example E (NCERT Example 15, p. 247). Differentiate f(x) = (x + 1)/x.

Step 1 — quotient rule: u = x + 1, v = x, u′ = 1, v′ = 1. Step 2 — apply formula: (1·x − (x+1)·1)/x². Step 3 — simplify: (x − x − 1)/x² = −1/x².

2.4 Common confusions / NTA trap points

• A function may be defined at x = a while lim_{x→a} f(x) does not exist (e.g. Fig 12.3, Fig 12.6) — students wrongly assume lim = f(a) always (NCERT §12.3 and Summary point 3, p. 221, 254).
• Conversely, the limit may exist where the function is not defined, e.g. (x² − 4)/(x − 2) at x = 2 (NCERT §12.3 illustration, p. 220).
• For a rational function with 0/0 form, you must factor and cancel before substituting; substituting blindly gives an undefined expression even though the limit may be a finite number or zero (NCERT §12.3.2, Example 2(ii)–(iv), p. 230–231).
• Theorem 2 lim (xⁿ − aⁿ)/(x − a) = n·a^(n−1) often hides under substitution: students forget to put it in the form f(y) where y → 1 (e.g. (1+x)^(1/2) − 1)/x via y = 1 + x, NCERT Example 3(ii), p. 233).
• The sin x / x and (1 − cos x)/x limits are about angles in radians only; using degrees is a classic trap (NCERT §12.4 proof, p. 234–236).
• Product rule (uv)′ = u′v + uv′ is not u′v′ — a frequent NTA distractor (NCERT §12.5.1, p. 244).
• Quotient rule has the sign and order fixed: numerator is u′v − uv′, not uv′ − u′v; reversing the order changes the sign (NCERT §12.5.1, p. 244–245).
• d/dx (cos x) = − sin x carries a minus sign that students drop (NCERT Summary, p. 255).
• Confusing the derivative as a number (f′(a)) with the derivative as a function (f′(x)); the first depends on a specific point, the second is itself a function of x.
• Forgetting to take limit before substitution in indeterminate forms 0/0; direct substitution gives "undefined", not the limit value.
• Misreading "instantaneous velocity" as average velocity over a finite interval; the limit gives the instantaneous (point-in-time) velocity.
• Mistakenly using (xⁿ − aⁿ)/(x − a) → n·aⁿ instead of n·a^(n−1); the exponent drops by one when the derivative is taken.