Q74. lim ex2−cos x→0 sin2 x (1) 2 (2) 32 (3) 5 (4) 3 4
What This Question Tests
This question evaluates the ability to calculate limits of indeterminate forms, which can be efficiently solved using L'Hopital's Rule or by applying series expansions of elementary functions around x=0.
Concepts Tested
Formulas Used
lim (f(x)/g(x)) = lim (f'(x)/g'(x)) for 0/0 form
Standard limits like lim (sin x)/x = 1
e^x ≈ 1+x+x²/2
cos x ≈ 1-x²/2
sin x ≈ x
📚 NCERT Sections This Tests
1.3 — Define The Following Terms:
Chemistry Class 11 · Chapter 1
1.3 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.
12.5 — A Hydrogen Atom Initially In The Ground Level Absorbs A Photon,
Physics Class 12 · Chapter 12
12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
14.2 — Which Of The Statements Given In Exercise 14.1 Is True For P-Type
Physics Class 12 · Chapter 14
14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
📋 Question Details
- Chapter
- Limits & Continuity
- Topic
- Limits using L'Hopital's Rule or series expansion
- Year
- 2015
- Shift
- 10 Apr Online
- Q Number
- Q74
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 5: Continuity and Differentiability
More from this Chapter
Q97.The function f : R ∼{0} →R given by f(x) = x1 − e2x−12 can be made continuous at x = 0 by defining f(0) as (1) 2 (2) −1 (3) 0 (4) 1
Q92.Let f(x) = −1) sin ( {(x0, if x = 1 JEE Main 2008 JEE Main Previous Year Paper (1) f is neither differentiable at x = 0 nor at x = 1 (2) f is differentiable at x = 0 and at x = 1 (3) f is differentiable at x = 0 but not at x = 1 (4) f is differentiable at x = 1 but not at x = 0
Q70.Let f : R →R be a positive increasing function with limx→∞ f(3x)f(x) = 1. Then limx→∞ f(2x)f(x) (1) 2 (2) 3 3 2 (3) 3 (4) 1
Q71.Consider the following statements P : Suman is brilliant Q : Suman is rich R : Suman is honest The negation of the statement "Suman is brilliant and dishonest if and only if Suman is rich" can be expressed as (1) ∼(Q ↔(P∧∼R)) (2) ∼Q ↔∼P ∧R (3) ∼(P∧∼R) ↔Q (4) ∼P ∧(Q ↔∼R)