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Limits and Derivatives — Class 11 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Limits and Derivatives" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Limits and Derivatives" — 8 important questions with detailed answers for CBSE boa…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

  1. Find the limit: lim(x→2) (x² - 4)/(x - 2)
  2. Find the limit: lim(x→0) (sin x)/x
  3. Find the derivative of f(x) = 3x² + 2x - 5 using the first principle of deriv…
  4. Find the derivative of f(x) = √(2x + 3)
  5. Find the derivative of f(x) = x² sin x
  6. Find the derivative of f(x) = (x² + 1)/(x - 1)
  7. + 2 more questions in the full chapter

Solutions Summary:

Question Status
Find the limit: lim(x→2) (x² - 4)/(x - 2) ✓ Solved
Find the limit: lim(x→0) (sin x)/x ✓ Solved
Find the derivative of f(x) = 3x² + 2x - 5 using the firs… ✓ Solved
Find the derivative of f(x) = √(2x + 3) ✓ Solved
Find the derivative of f(x) = x² sin x ✓ Solved
Find the derivative of f(x) = (x² + 1)/(x - 1) ✓ Solved

Showing 6 of 8 questions

Q1: Find the limit: lim(x→2) (x² - 4)/(x - 2)

Find: lim(x→2) (x² - 4)/(x - 2) Step 1: Check for direct substitution: When x = 2: (2² - 4)/(2 - 2) = 0/0 (indeterminate form) Step 2: Factorize the numerator: x² - 4 = (x + 2)(x - 2) Step 3: Cancel common factors: lim(x→2) (x² - 4)/(x - 2) = lim(x→2) [(x + 2)(x - 2)]/(x - 2) = lim(x→2) (x + 2) Step 4: Apply direct substitution: = 2 + 2 = 4 Final Answer: The limit is 4

Q2: Find the limit: lim(x→0) (sin x)/x

Find: lim(x→0) (sin x)/x Step 1: This is a standard limit form. Step 2: Direct substitution gives 0/0 (indeterminate). Step 3: Use L'Hôpital's Rule: lim(x→0) (sin x)/x = lim(x→0) (d/dx sin x)/(d/dx x) = lim(x→0) (cos x)/1 = cos 0 = 1 Alternatively, this is a well-known standard limit: lim(x→0) (sin x)/x = 1 Final Answer: The limit is 1

Q3: Find the derivative of f(x) = 3x² + 2x - 5 using the first principle of derivatives.

Find the derivative of f(x) = 3x² + 2x - 5 using first principle. Step 1: Use the definition: f'(x) = lim(h→0) [f(x + h) - f(x)]/h Step 2: Calculate f(x + h): f(x + h) = 3(x + h)² + 2(x + h) - 5 = 3(x² + 2xh + h²) + 2x + 2h - 5 = 3x² + 6xh + 3h² + 2x + 2h - 5 Step 3: Calculate f(x + h) - f(x): f(x + h) - f(x) = [3x² + 6xh + 3h² + 2x + 2h - 5] - [3x² + 2x - 5] = 6xh + 3h² + 2h = h(6x + 3h + 2) Step 4: Apply the limit: f'(x) = lim(h→0) [h(6x + 3h + 2)]/h = lim(h→0) (6x + 3h + 2) = 6x + 0 + 2 =...

Q4: Find the derivative of f(x) = √(2x + 3)

Find the derivative of f(x) = √(2x + 3) Step 1: Rewrite the function: f(x) = (2x + 3)^(1/2) Step 2: Use the chain rule: f'(x) = d/dx[(2x + 3)^(1/2)] = (1/2)(2x + 3)^(-1/2) × d/dx(2x + 3) Step 3: Find the derivative of the inner function: d/dx(2x + 3) = 2 Step 4: Substitute: f'(x) = (1/2)(2x + 3)^(-1/2) × 2 = (2x + 3)^(-1/2) = 1/√(2x + 3) Final Answer: f'(x) = 1/√(2x + 3)

Q5: Find the derivative of f(x) = x² sin x

Find the derivative of f(x) = x² sin x Step 1: This is a product of two functions. Let u = x² and v = sin x Step 2: Use the product rule: f'(x) = u'v + uv' Step 3: Find u' and v': u' = 2x v' = cos x Step 4: Apply the product rule: f'(x) = (2x)(sin x) + (x²)(cos x) = 2x sin x + x² cos x = x(2 sin x + x cos x) Final Answer: f'(x) = 2x sin x + x² cos x

Q6: Find the derivative of f(x) = (x² + 1)/(x - 1)

Find the derivative of f(x) = (x² + 1)/(x - 1) Step 1: This is a quotient of two functions. Let u = x² + 1 and v = x - 1 Step 2: Use the quotient rule: f'(x) = [u'v - uv']/v² Step 3: Find u' and v': u' = 2x v' = 1 Step 4: Apply the quotient rule: f'(x) = [(2x)(x - 1) - (x² + 1)(1)]/(x - 1)² = [2x² - 2x - x² - 1]/(x - 1)² = [x² - 2x - 1]/(x - 1)² Final Answer: f'(x) = (x² - 2x - 1)/(x - 1)²

Showing 6 of 8 questions. Visit the full page for complete solutions.

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