Example 3 - Chapter 13 Class 11 Limits and Derivatives - Part 5

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Example 3 - Chapter 13 Class 11 Limits and Derivatives - Part 6

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Example 3 - Chapter 13 Class 11 Limits and Derivatives - Part 7

  1. Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
  2. Serial order wise

Transcript

Example 3 Evaluate: (ii) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’0) (โˆš(1 + ๐‘ฅ) โˆ’ 1)/๐‘ฅ (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’0) (โˆš(1 + x )โˆ’ 1)/x Putting x = 0 = (โˆš(1 + 0) โˆ’ 1)/0 = (โˆš(1 ) โˆ’ 1)/0 = (1 โˆ’ 1)/0 = 0/0 Since it is a 0/0 form We simplify the equation Putting y = 1 + x โ‡’ y โ€“ 1 = x As x โ†’ 0 y โ†’ 1 + 0 y โ†’ 1 So, our equation becomes (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’0) (โˆš(1 + ๐‘ฅ )โˆ’ 1)/๐‘ฅ = (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฆโ†’1) (โˆš๐‘ฆ โˆ’ 1)/(๐‘ฆ โˆ’ 1) = (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) ( ๐‘ฆ^((โˆ’1)/2) โˆ’ 1)/(๐‘ฆ โˆ’ 1) = (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) ( ๐‘ฆ^((โˆ’1)/2) โˆ’ 1^((โˆ’1)/2))/(๐‘ฆ โˆ’ 1) = 1/2 ร— 1^((โˆ’1)/2 โˆ’ 1) = 1/2 ร— 1 = ๐Ÿ/๐Ÿ

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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.