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Test: Logarithmic And Exponential Limits - Commerce MCQ


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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
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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
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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

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Test: Logarithmic And Exponential Limits - Question 3
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Test: Logarithmic And Exponential Limits - Question 4
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Test: Logarithmic And Exponential Limits - Question 5
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Test: Logarithmic And Exponential Limits - Question 6
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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
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Test: Logarithmic And Exponential Limits - Question 8
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 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
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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
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