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Example 4 (i) - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
Last updated at Aug. 28, 2021 by Teachoo
Transcript
Example, 4 Evaluate: (i) limx→0 sin4xsin 2x limx→0 sin4xsin 2x = limx→0 sin 4x × limx→0 1 sin2𝑥 Multiplying & dividing by 4x = limx→0 sin 4x . 4𝑥4𝑥 × limx→0 1 sin2𝑥 = limx→0 sin4𝑥4𝑥 . 4x × limx→0 1 sin2𝑥 = 𝐥𝐢𝐦𝐱→𝟎 𝒔𝒊𝒏𝟒𝒙𝟒𝒙 × limx→0 4x × limx→0 1 sin2𝑥 = 1 × limx→0 4x × limx→0 1 sin2𝑥 = 1 × limx→0 4x × limx→0 1 sin2𝑥 = limx→0 1 sin2𝑥 . 4x Multiplying & dividing by 2x = limx→0 1sin 2x × 4x × 2𝑥2𝑥 = limx→0 2𝑥sin 2x × 4𝑥2𝑥 = limx→0 2𝑥 sin2𝑥 × 2 = 2 limx→0 2𝑥 sin2𝑥 = 2 ÷ 𝐥𝐢𝐦𝐱→𝟎 𝐬𝐢𝐧𝟐𝒙𝟐𝒙 = 2 ÷ 1 = 21 = 2
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Examples (Term 1 and Term 2)
Example 1 (i)
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii) Important
Example 2 (iii) Important
Example 2 (iv)
Example 2 (v)
Example 3 (i) Important
Example 3 (ii) Important
Example 4 (i) You are here
Example 4 (ii) Important
Example 5
Example 6
Example 7 Important
Example 8
Example 9
Example 10 Important
Example 11
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16
Example 17 Important
Example 18
Example 19 (i) Important
Example 19 (ii)
Example 20 (i)
Example 20 (ii) Important
Example 21 (i)
Example 21 (ii) Important
Example 22 (i)
Example 22 (ii) Important
Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
Serial order wise
- Ex 13.1 (Term 1)
- Ex 13.2 (Term 2)
-
Examples (Term 1 and Term 2)
- Miscellaneous (Term 1 and Term 2)
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.