# Example 4 - Chapter 13 Class 11 Limits and Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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 Example 4 Evaluate: (ii) limx→0 tanxx limx→0 tanxx = limx→0 1𝑥 × tan x = limx→0 1𝑥 × sinx cosx = limx→0 sinxx × 1 cos𝑥 = 𝐥𝐢𝐦𝐱→𝟎 𝐬𝐢𝐧𝐱𝐱 × limx→0 1cos x = 1 × limx→0 1cos x = 1 × limx→0 1cos x = limx→0 1cos x Putting x = 0 = 1 cos0 = 11 = 1

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.