Q75.If x→∞(1lim + (1) 2 (2) 32 (3) 1 (4) 2 2 3
What This Question Tests
The question tests the ability to evaluate a limit of the form 1^infinity using the standard formula and algebraic manipulation to compare it with a given exponential value.
Concepts Tested
Formulas Used
lim (1 + f(x))^(g(x)) = e^(lim f(x)g(x)) as f(x) -> 0
📚 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.
11.3 — The Photoelectric Cut-Off Voltage In A Certain Experiment Is 1.5 V.
Physics Class 12 · Chapter 11
11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
1.1 — Define The Term Solution. How Many Types Of Solutions Are Formed? Write Briefly
Chemistry Class 11 · Chapter 1
1.1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
📋 Question Details
- Chapter
- Limits & Continuity
- Topic
- Limits of the form 1^infinity
- Year
- 2016
- Shift
- 09 Apr Online
- Q Number
- Q75
- 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)