🎯 Key Points
- Standard limits to memorize: lim(x→0) sinx/x = 1, lim(x→0)(1-cosx)/x = 0, lim(x→∞)(1+1/x)ˣ = e — these appear disguised inside countless other limit problems
- Derivative = slope of the TANGENT at a point = limiting slope of the secant as the second point approaches the first (the geometric meaning behind the algebra)
- Product rule (fg)'=f'g+fg' vs Quotient rule (f/g)'=(f'g-fg')/g² — note the quotient rule's numerator order matters (f'g MINUS fg', not the reverse)
- Critical point test: f'(x)=0 locates candidates; f''(x)>0 → local minimum, f''(x)<0 → local maximum (second derivative test)
Limits & Derivatives
Calculus is the study of continuous change. Differential calculus finds rates of change (derivatives); integral calculus finds areas/accumulations.
Limits
As point Q slides along the curve toward P, the secant line PQ rotates into the tangent line at P, whose slope is the derivative.
- lim(x→a) f(x) is the value f approaches as x approaches a
- Standard limits: lim(x→0) sinx/x = 1; lim(x→0) (1-cosx)/x = 0; lim(x→∞)(1+1/x)ˣ = e
- L'Hopital's rule: if 0/0 or ∞/∞ form, lim f/g = lim f'/g'
Differentiation Rules
- Power rule: d/dx(xⁿ) = nxⁿ⁻¹
- Constant: d/dx(c) = 0
- Sum: d/dx(f+g) = f' + g'
- Product: d/dx(fg) = f'g + fg'
- Quotient: d/dx(f/g) = (f'g - fg')/g²
- Chain rule: d/dx[f(g(x))] = f'(g(x)) × g'(x)
Standard Derivatives
- d/dx(sinx) = cosx
- d/dx(cosx) = -sinx
- d/dx(tanx) = sec²x
- d/dx(eˣ) = eˣ
- d/dx(ln x) = 1/x
- d/dx(aˣ) = aˣ ln a
Applications
- Finding maxima/minima: f'(x) = 0 gives critical points; f''(x) > 0 is minimum, < 0 is maximum
- Rate of change: velocity = dx/dt, acceleration = d²x/dt²
- Tangent at (a,b): y - b = f'(a)(x - a)
- Normal: y - b = -1/f'(a) × (x - a)
Algebra of Limits
If lim(x→a) f(x) = L and lim(x→a) g(x) = M both exist, then limits distribute over the basic operations:
- Sum/difference: lim [f(x) ± g(x)] = L ± M
- Product: lim [f(x)·g(x)] = L·M; and lim [k·f(x)] = k·L for a constant k
- Quotient: lim [f(x)/g(x)] = L/M, provided M ≠ 0
- Power: lim [f(x)]ⁿ = Lⁿ
These rules let a complicated limit be broken into simpler pieces, each evaluated separately and then recombined.
Limits of Polynomials and Rational Functions
- Polynomials: a polynomial is continuous everywhere, so lim(x→a) p(x) = p(a) — just substitute x = a directly.
- Rational functions p(x)/q(x): if q(a) ≠ 0, substitute to get p(a)/q(a).
- 0/0 form: if both p(a) = 0 and q(a) = 0, then (x − a) is a common factor. Cancel it (or use the standard result below), then substitute. Example: lim(x→2)(x²−4)/(x−2) = lim(x→2)(x+2) = 4.
Some Standard Algebraic and Exponential Limits
Besides the trigonometric limits above, these results are used constantly and are worth memorising:
- lim(x→a) (xⁿ − aⁿ)/(x − a) = n·aⁿ⁻¹ (valid for all rational n)
- lim(x→0) (aˣ − 1)/x = ln a (for a > 0)
- lim(x→0) (eˣ − 1)/x = 1 (the special case a = e)
- lim(x→0) ln(1 + x)/x = 1
- lim(x→0) tan x / x = 1
To use them, algebraically massage the expression so the target sub-expression matches one of these standard forms (e.g. multiply and divide by a suitable variable).
The Derivative as a Limit (First Principle)
The derivative of f at x is defined as the limit of the average rate of change over a shrinking interval:
f′(x) = lim(h→0) [f(x + h) − f(x)] / h
- The quotient [f(x+h) − f(x)]/h is the slope of the secant through (x, f(x)) and (x+h, f(x+h)); as h→0 it becomes the slope of the tangent.
- The derivative at a point a is f′(a) = lim(h→0) [f(a+h) − f(a)]/h, a single number giving the instantaneous rate of change at x = a.
- If this limit fails to exist (e.g. a sharp corner like f(x) = |x| at x = 0), the function is not differentiable there.
- Differentiability at a point always implies continuity there, but continuity does not guarantee differentiability.
🚀 JEE Advanced Edge
First-principles (limit definition) differentiation: f'(x) = lim(h→0) [f(x+h)-f(x)]/h is the rigorous definition underlying every shortcut rule (power, product, chain) — JEE occasionally demands proving a derivative "from first principles" specifically to test whether the limit definition is understood, not just the memorized shortcut formulas.
Why L'Hopital's rule can be applied repeatedly: If after one application of L'Hopital's rule the limit is STILL in 0/0 or ∞/∞ form, the rule can be reapplied (differentiating numerator and denominator again) — but each application must be re-verified to actually be in an indeterminate form first, since applying it to a non-indeterminate expression gives a wrong answer.
Worked problem: Evaluate lim(x→0) (1-cos4x)/x². Approach: Rewrite 1-cos4x = 2sin²(2x) (using 1-cosθ=2sin²(θ/2) with θ=4x). So the limit becomes lim(x→0) 2sin²(2x)/x² = 2·lim(x→0)[sin(2x)/x]² = 2·[lim(x→0) (sin2x/2x)·2]² = 2·(1·2)² = 2×4 = 8.
Worked Example: Limit Using Standard Results
Evaluate lim(x→0) (e^(3x) − 1) / (sin 2x).
Rewrite using standard limits lim(u→0)(eᵘ−1)/u = 1 and lim(u→0)(sinu)/u = 1:
= lim(x→0) [(e^(3x)−1)/(3x)] · 3 / [(sin2x)/(2x)] · 2 = (1 × 3) / (1 × 2) = 3/2. Multiply and divide by the "standard-limit denominators" (3x and 2x) to reduce every limit to a known form.
Worked Example: Derivative by First Principles
Find the derivative of f(x) = x² + 3x from first principles.
f′(x) = lim(h→0) [f(x+h) − f(x)] / h = lim(h→0) [(x+h)² + 3(x+h) − x² − 3x] / h = lim(h→0) [2xh + h² + 3h] / h.
= lim(h→0) (2x + h + 3) = 2x + 3. First-principles questions follow this four-step pattern: expand, simplify, cancel h, then take the limit.