Commerce Exam  >  Commerce Tests  >  Mathematics (Maths) Class 11  >  Test: Logarithmic And Exponential Limits - Commerce MCQ

Test: Logarithmic And Exponential Limits - Commerce MCQ


Test Description

10 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Logarithmic And Exponential Limits

Test: Logarithmic And Exponential Limits for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Logarithmic And Exponential Limits questions and answers have been prepared according to the Commerce exam syllabus.The Test: Logarithmic And Exponential Limits MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Logarithmic And Exponential Limits below.
Solutions of Test: Logarithmic And Exponential Limits questions in English are available as part of our Mathematics (Maths) Class 11 for Commerce & Test: Logarithmic And Exponential Limits solutions in Hindi for Mathematics (Maths) Class 11 course. Download more important topics, notes, lectures and mock test series for Commerce Exam by signing up for free. Attempt Test: Logarithmic And Exponential Limits | 10 questions in 10 minutes | Mock test for Commerce preparation | Free important questions MCQ to study Mathematics (Maths) Class 11 for Commerce Exam | Download free PDF with solutions
Test: Logarithmic And Exponential Limits - Question 1

Detailed Solution for Test: Logarithmic And Exponential Limits - Question 1

lim(x → 0) (ex + e-x - 2)/x2
it is (0/0) form, hence apply L hospital rule
Taking derivative, we get
lim(x → 0) (ex - e-x)/2x
lim(x → 0) (ex + e-x)/2
= (1 + 1)/2 
⇒ 2/2 = 1

Test: Logarithmic And Exponential Limits - Question 2

Detailed Solution for Test: Logarithmic And Exponential Limits - Question 2

 lim(x → 0) log(1+x)/[√(1+x)-1)]
= lim(x → 0) log(1+x)/[√(1+x)-1)] * {[√(1+x)+1) * [√(1+x)+1)]}
= lim(x → 0) [log(1+x)/x] * (√(1+x)+1)
= 1 * (2)
= 2

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Logarithmic And Exponential Limits - Question 3

Test: Logarithmic And Exponential Limits - Question 4

Test: Logarithmic And Exponential Limits - Question 5

Test: Logarithmic And Exponential Limits - Question 6

Detailed Solution for Test: Logarithmic And Exponential Limits - Question 6

lim(x → 0) (x2x - x)/(1-cosx)
lim(x → 0) x2(2x - 1)/x(1-cosx)
By formula : lim(x → 0) (ax - 1)/x = log a
lim(x → 0) x2 log2/(1-cosx)
Differentiate it 
lim(x → 0) 2x log2/(sinx)
= 2 log2

Test: Logarithmic And Exponential Limits - Question 7

Test: Logarithmic And Exponential Limits - Question 8

 is equal to

Detailed Solution for Test: Logarithmic And Exponential Limits - Question 8


⇒ Using L'Hospital,

= 1/2

Test: Logarithmic And Exponential Limits - Question 9

Find log5125

Detailed Solution for Test: Logarithmic And Exponential Limits - Question 9

53 = 125 
As we know, 
ab = c ⇒ logac = b 
Therefore; Log5125 = 3

Test: Logarithmic And Exponential Limits - Question 10

75 videos|238 docs|91 tests
Information about Test: Logarithmic And Exponential Limits Page
In this test you can find the Exam questions for Test: Logarithmic And Exponential Limits solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Logarithmic And Exponential Limits, EduRev gives you an ample number of Online tests for practice

Up next

75 videos|238 docs|91 tests
Download as PDF

Up next