Test: Logarithmic And Exponential Limits - Commerce MCQ

# 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.
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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

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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

## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests
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## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests