Take timed practice tests on Limits and Continuity for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus one 60-question module test. Each question is original and calibrated from the uploaded material pattern without copying PDF wording.
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1. lim(x→2) (x² − 4)/(x − 2) equals:
Explanation: Factor: (x²−4)/(x−2) = (x+2)(x−2)/(x−2) = x+2 (for x ≠ 2). As x → 2: 2 + 2 = 4.
2. lim(x→0) (sin x)/x equals:
Explanation: This is a standard limit: lim(x→0) (sin x)/x = 1. It can be proved using the squeeze theorem.
3. lim(x→0) (1 + x)^(1/x) equals:
Explanation: This is the standard definition of e: lim(x→0) (1+x)^(1/x) = e ≈ 2.718.
4. A function f(x) is continuous at x = a if:
Explanation: Continuity at x = a requires three conditions: f(a) is defined, lim(x→a) f(x) exists, and the limit equals f(a).
5. If f(x) = |x|/x for x ≠ 0, then lim(x→0⁺) f(x) is:
Explanation: For x > 0: |x| = x, so f(x) = x/x = 1. As x → 0⁺, f(x) → 1.
6. lim(x→∞) (3x + 1)/(2x − 5) equals:
Explanation: Divide numerator and denominator by x: lim (3 + 1/x)/(2 − 5/x) = (3 + 0)/(2 − 0) = 3/2.
7. lim(x→3) (x³ − 27)/(x − 3) equals:
Explanation: Using the identity aⁿ − bⁿ = (a−b)(aⁿ⁻¹ + aⁿ⁻²b + ... + bⁿ⁻¹): x³ − 27 = (x−3)(x² + 3x + 9). Cancel (x−3): limit = x² + 3x + 9 at x = 3 = 9 + 9 + 9 = 27.
8. lim(x→0) (tan x)/x equals:
Explanation: lim(x→0) (tan x)/x = lim (sin x)/(x cos x) = lim(sin x/x) × lim(1/cos x) = 1 × 1 = 1.
9. lim(x→0) (1 − cos x)/x² equals:
Explanation: Using half-angle: 1 − cos x = 2sin²(x/2). So (1−cos x)/x² = 2sin²(x/2)/x² = (1/2)[sin(x/2)/(x/2)]² → 1/2 as x → 0.
10. lim(x→0) (eˣ − 1)/x equals:
Explanation: Standard limit: lim(x→0) (eˣ−1)/x = 1. This follows from the derivative of eˣ at x = 0 being 1.
11. lim(x→0) (sin 3x)/(sin 5x) equals:
Explanation: (sin 3x)/(sin 5x) = [(sin 3x)/(3x)] × 3x × (5x)/[(sin 5x)/(5x)] × 1/(5x) = (3/5) × [sin 3x / 3x] / [sin 5x / 5x] → 3/5 × 1/1 = 3/5.
12. f(x) = (x² − 1)/(x − 1) for x ≠ 1 and f(1) = 3. Is f continuous at x = 1?
Explanation: lim(x→1) (x²−1)/(x−1) = lim(x+1) = 2. But f(1) = 3. Since limit ≠ f(1), f is not continuous at x = 1.
13. The function f(x) = sin x / x at x = 0 has:
Explanation: lim(x→0) sin x / x = 1 exists, but the function is not defined at x = 0. If we define f(0) = 1, f becomes continuous — so it has a removable discontinuity.
14. lim(x→0) [x]/x (where [x] is the greatest integer function) is:
Explanation: Right limit: as x → 0⁺, [x] = 0, so [x]/x = 0/x → 0. Left limit: as x → 0⁻, [x] = −1, so [x]/x = −1/x → +∞. Limits differ, so the limit does not exist.
15. lim(x→0) (aˣ − 1)/x equals:
Explanation: Using the derivative definition: d/dx(aˣ) at x = 0 = aˣ ln a at x = 0 = ln a. Equivalently: aˣ = e^(x ln a) so (aˣ−1)/x = (e^(x ln a)−1)/x → ln a as x → 0.
16. lim(x→0⁺) x^x equals:
Explanation: Let L = lim(x→0⁺) x^x. Take ln: ln L = lim(x→0⁺) x ln x. As x → 0⁺, x → 0 and ln x → −∞, but x ln x → 0 (exponential dominates). So ln L = 0 → L = e⁰ = 1.
17. If f(x) = x² sin(1/x) for x ≠ 0 and f(0) = 0, then f is:
Explanation: Continuity: |f(x)| = |x² sin(1/x)| ≤ x² → 0 as x → 0 = f(0) ✓. Differentiability: f'(0) = lim(x→0) f(x)/x = lim x sin(1/x). Since |x sin(1/x)| ≤ |x| → 0, f'(0) = 0 ✓.
18. lim(x→1) (xⁿ − 1)/(x − 1) equals:
Explanation: Using the standard result: lim(x→a) (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹. Here a = 1: lim(x→1) (xⁿ − 1)/(x − 1) = n × 1ⁿ⁻¹ = n.
19. lim(x→π/2) (1 − sin x)/cos²x equals:
Explanation: cos²x = 1 − sin²x = (1−sin x)(1+sin x). So (1−sin x)/cos²x = 1/(1+sin x). At x = π/2: 1/(1+1) = 1/2.
20. lim(n→∞) (1/n²)(1 + 2 + 3 + ... + n) equals:
Explanation: Sum = n(n+1)/2. Divided by n²: (n+1)/(2n) = (1 + 1/n)/2 → 1/2 as n → ∞.