Quick Summary
Limits describe the behaviour of a function near a point, without necessarily reaching it. Continuity formalises the intuitive idea of an unbroken graph. Together they are the foundation of differential and integral calculus. Every derivative definition uses a limit.
Concept of a Limit
lim(x→a) f(x) = L means f(x) → L as x → a (from both sides).
- Left-hand limit (LHL): lim(x→a⁻) f(x)
- Right-hand limit (RHL): lim(x→a⁺) f(x)
Limit exists ↔ LHL = RHL = L
Standard Limits
| Limit | Value |
|---|---|
| lim(x→0) sinx/x | 1 |
| lim(x→0) tanx/x | 1 |
| lim(x→0) (1−cosx)/x² | 1/2 |
| lim(x→0) (eˣ−1)/x | 1 |
| lim(x→0) (aˣ−1)/x | ln a |
| lim(x→0) (1+x)^(1/x) | e |
| lim(x→∞) (1+1/x)ˣ | e |
| lim(x→a) (xⁿ−aⁿ)/(x−a) | naⁿ⁻¹ |
| lim(x→0) log(1+x)/x | 1 |
Limit Laws
For lim(x→a) f(x) = L and lim(x→a) g(x) = M:
| Law | Result |
|---|---|
| Sum | lim [f+g] = L + M |
| Product | lim [f·g] = L × M |
| Quotient | lim [f/g] = L/M (M ≠ 0) |
| Scalar | lim [k·f] = k·L |
Evaluating Limits
Direct substitution: If f(a) gives a finite value, lim(x→a) f(x) = f(a).
Factorisation: Use when substitution gives 0/0 — factorise and cancel.
Rationalisation: Multiply numerator and denominator by the conjugate.
Standard limit substitution: Transform the expression to match a standard form.
L'Hôpital's Rule (for 0/0 or ∞/∞ forms): lim(x→a) f(x)/g(x) = lim(x→a) f'(x)/g'(x)
Continuity
f(x) is continuous at x = a if: 1. f(a) is defined 2. lim(x→a) f(x) exists 3. lim(x→a) f(x) = f(a)
Continuous on [a,b]: Continuous at every point in the interval.
Types of Discontinuity
| Type | Condition |
|---|---|
| Removable | Limit exists but ≠ f(a), or f(a) undefined |
| Jump | LHL ≠ RHL (both finite) |
| Infinite | Limit → ∞ (vertical asymptote) |
Key Theorems
- Intermediate Value Theorem (IVT): If f is continuous on [a,b] and f(a) < k < f(b), then ∃ c ∈ (a,b) with f(c) = k.
- All polynomials, trigonometric, exponential, and logarithmic functions are continuous on their domains.
Worked Examples
Example 1: Evaluate lim(x→2) (x²−4)/(x−2). = lim(x→2) (x+2)(x−2)/(x−2) = lim(x→2) (x+2) = 4
Example 2: Evaluate lim(x→0) sin(3x)/x. = lim(x→0) 3 × sin(3x)/(3x) = 3 × 1 = 3
Example 3: Evaluate lim(x→0) (√(1+x) − 1)/x. Multiply by (√(1+x)+1): = lim(x→0) x / (x(√(1+x)+1)) = 1/(1+1) = 1/2
Example 4: Check continuity of f(x) = x² at x = 2. f(2) = 4. lim(x→2) x² = 4. Since f(2) = limit = 4, continuous at x = 2.
Important Questions
One Mark
- Evaluate lim(x→3) (x² − 9)/(x − 3).
- State the condition for a function to be continuous at a point.
- Find lim(x→0) sin(5x)/x.
- Is f(x) = |x| continuous at x = 0? Why?
Short Answer (3 marks)
- Evaluate lim(x→0) (1 − cos 2x)/x².
- Evaluate lim(x→2) (x³ − 8)/(x² − 4).
- Find the value of k so that f(x) = kx + 1 for x ≤ 2, and 3x − 1 for x > 2 is continuous at x = 2.
Long Answer (5 marks)
- Prove that lim(x→0) sinx/x = 1 using the squeeze theorem (geometric proof).
- Find all points of discontinuity of f(x) = [x] (greatest integer function) and classify them.
Common Mistakes
- Applying direct substitution when the result is 0/0 — always factor or rationalise first.
- Using L'Hôpital's Rule on forms other than 0/0 or ∞/∞ — check the indeterminate form first.
- Continuity: checking only lim = f(a) but not verifying f(a) is defined.
- Standard limits: misapplying lim sin x/x = 1 — valid only as x→0, not for all x.
Board Exam Tips
- "Evaluate the limit" problems of the 0/0 factorisation type appear every year as 2–3 mark questions.
- "Find k for continuity" problems (piecewise functions) are standard 3-mark questions — equate LHL = RHL = f(a).
- The proof of lim sinx/x = 1 is a textbook 5-mark derivation — prepare it fully.
- JEE: Squeeze theorem, limits of composite functions, and one-sided limits are tested regularly.