Differential Calculus (Limits, Continuity & AOD)
Practice JEE Main Limits, Continuity, Differentiability & AOD PYQs. Master L’Hôpital’s rule, 1^∞ shortcut, Rolle’s theorem & maxima/minima.
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4-Step Solution & Tricks
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Differential Calculus (Limits, Continuity & AOD):
- Evaluation of Limits & L’Hôpital’s Rule
- 1^∞ Indeterminate Form Shortcut: e^(lim (f - 1)·g)
- Continuity & Differentiability at a Point
- Rolle’s Theorem & Lagrange’s Mean Value Theorem (LMVT)
- Monotonicity, Maxima & Minima, Tangents & Normals
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Differential Calculus questions in under 45 seconds:
1^∞ Form: lim [f(x)]^(g(x)) = e^(lim [f(x) - 1] · g(x)).
Standard Limit: lim (sin x)/x = 1, lim (tan x)/x = 1, lim (eˣ - 1)/x = 1.
LMVT: f’(c) = [f(b) - f(a)] / (b - a).
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: 2
- Step 1: Rewrite standard trigonometric limits as x → 0: (1 - cos 2x) = 2 sin² x.
- Step 2: Expression becomes lim (x → 0) [2 sin² x · (3 + cos x)] / (x · tan 4x).
- Step 3: Group standard limits: 2 · (sin x / x)² · (3 + cos x) · (4x / tan 4x) · (x² / (x · 4x)).
- Step 4: Notice x² / (x · 4x) = 1/4. Then substitute x = 0: 2 · (1)² · (3 + 1) · (1) · (1/4) = 2 · 4 · (1/4) = 2.
⚡ Speed Tip: Trig Limit Shortcut: (1 - cos 2x) ≈ (2x)²/2 = 2x², and x tan 4x ≈ 4x². Limit = (2x² · 4)/(4x²) = 2.
💡 View Step-by-Step Solution & Answer
Correct Answer: 0
- Step 1: For x ∈ (1, 2), x - 1 > 0, so |x - 1| = x - 1.
- Step 2: For x ∈ (1, 2), x - 2 < 0, so |x - 2| = -(x - 2) = 2 - x.
- Step 3: Therefore, on the open interval (1, 2), f(x) = (x - 1) + (2 - x) = 1 (a constant function).
- Step 4: The derivative of a constant function is zero: f'(x) = d/dx(1) = 0. Specifically, f'(1.5) = 0.
⚡ Speed Tip: Plateau Principle: Between the roots of |x - a| + |x - b|, the sum is constant (b - a = 1), so the slope is 0.
💡 View Step-by-Step Solution & Answer
Correct Answer: -1/3
- Step 1: Differentiate y with respect to x: dy/dx = d/dx(2x² + 3 sin x) = 4x + 3 cos x.
- Step 2: Evaluate the tangent slope at x = 0: m_tangent = (dy/dx)|_{x=0} = 4(0) + 3 cos(0) = 0 + 3(1) = 3.
- Step 3: The normal line is perpendicular to the tangent line: m_normal · m_tangent = -1.
- Step 4: Therefore, m_normal = -1 / m_tangent = -1 / 3.
⚡ Speed Tip: Normal Slope: m_normal = -1 / (dy/dx). 4(0) + 3 cos(0) = 3 ⇒ m_normal = -1/3.
❓ Frequently Asked Questions (FAQs)
What is the fastest way to solve 1^∞ limits in JEE Main?
Convert directly using the formula e^(lim (f(x) - 1) · g(x)). This converts the power into a simple product limit.