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JEE Main Unit 1B1: Limit Concepts and Graphical Limits — practice questions

Explore 25 free Unit 1B1: Limit Concepts and Graphical Limits questions in JEE Main Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 1B1: Limit Concepts and Graphical Limits questions and exercises with answers

Showing questions 1–25 of 25.

lim_(x→2) (x² - 4)/(x - 2) =Read the answer and worked solution →lim_(x→0) sin(x)/x =Read the answer and worked solution →For limit to EXIST at x = a, left-hand and right-hand limits must:Read the answer and worked solution →A function f is continuous at x = a if:Read the answer and worked solution →d/dx (xⁿ) = ?Read the answer and worked solution →lim_(x→0) (1 - cos x)/x² =Read the answer and worked solution →lim_(x→∞) (1 + 1/x)^x =Read the answer and worked solution →lim_(x→0) (e^x - 1)/x =Read the answer and worked solution →lim_(x→0) (sin 3x)/(sin 5x) =Read the answer and worked solution →d/dx (sin x) =Read the answer and worked solution →d/dx (eˣ) =Read the answer and worked solution →d/dx (ln x) =Read the answer and worked solution →Find d/dx (x² sin x) by product rule:Read the answer and worked solution →By chain rule, d/dx [sin(x²)] =Read the answer and worked solution →lim_(x→∞) (3x² - 5)/(7x² + 2) =Read the answer and worked solution →l'Hôpital's rule applies to limits of form:Read the answer and worked solution →Find dy/dx by implicit differentiation for x² + y² = 25:Read the answer and worked solution →d/dx (xˣ) where x > 0:Read the answer and worked solution →For f(x) = sin(1/x) for x ≠ 0, what is the limit as x → 0?Read the answer and worked solution →Differentiation by chain rule for d/dx [ln(sin x)]:Read the answer and worked solution →lim_(x→0) [(1 + x)^(1/x)] =Read the answer and worked solution →Higher-order derivative: d²/dx² (x⁴) =Read the answer and worked solution →f(x) is discontinuous at x = a if which condition fails?Read the answer and worked solution →For y = x³ - 6x² + 9x, find critical points (where y' = 0):Read the answer and worked solution →By Taylor expansion around x = 0: e^x = 1 + x + ___:Read the answer and worked solution →

Worked practice questions

lim_(x→2) (x² - 4)/(x - 2) =

  1. 4 (factor: x + 2 → 4 at x = 2)
  2. Infinity
  3. 2
  4. 0
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Correct answer: A. 4 (factor: x + 2 → 4 at x = 2)

x² - 4 = (x-2)(x+2). **Cancel (x-2):** limit is x + 2 = 4 at x = 2. (Indeterminate 0/0 resolved by factoring.)

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lim_(x→0) sin(x)/x =

  1. 1 (standard limit)
  2. x
  3. 0
  4. Infinity
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Correct answer: A. 1 (standard limit)

**Standard limit:** lim_(x→0) sin(x)/x = 1. Used throughout calculus. Proved via squeeze theorem or geometric arguments.

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For limit to EXIST at x = a, left-hand and right-hand limits must:

  1. Be different
  2. Be EQUAL
  3. Be zero
  4. Be infinity
Show answer and solution

Correct answer: B. Be EQUAL

Limit exists at x = a iff LHL = RHL (both must equal and be finite). If unequal or one is infinite, limit does not exist.

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