Example 3 - Evaluate (i) lim x->1 x15 - 1/x10 - 1 - Chapter 13

Example 3 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Example 3 - Chapter 13 Class 11 Limits and Derivatives - Part 3 Example 3 - Chapter 13 Class 11 Limits and Derivatives - Part 4

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Transcript

Example 3 Evaluate: (i) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) (๐‘ฅ 15 โˆ’ 1)/(๐‘ฅ10 โˆ’ 1) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) (๐‘ฅ 15 โˆ’ 1)/(๐‘ฅ10 โˆ’ 1) = (ใ€–(1)ใ€—^15 โˆ’ 1)/(ใ€–(1)ใ€—^10 โˆ’ 1) = (1 โˆ’ 1)/(1 โˆ’ 1) = 0/0 Since it is form 0/0, We can solve by using theorem (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’๐‘Ž) (๐‘ฅ^๐‘› โˆ’ ๐‘Ž^๐‘›)/(๐‘ฅ โˆ’ ๐‘Ž) = na n โ€“ 1 Hence, (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) (๐‘ฅ^15 โˆ’ 1)/(๐‘ฅ^10 โˆ’ 1) = (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) ๐‘ฅ^15 โ€“ 1 รทlimโ”ฌ(xโ†’1) x10 โ€“ 1 = (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) ๐‘ฅ^15 โ€“ ใ€–(1)ใ€—^15 รท limโ”ฌ(xโ†’1) x10 โ€“ (1)10 Multiplying and dividing by x โ€“ 1 = (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) (๐‘ฅ^15 โˆ’ 1^15)/(๐‘ฅ โˆ’ 1) รท (๐‘™๐‘–๐‘š)โ”ฌ(๐‘งโ†’1) (๐‘ฅ^10 โˆ’ ใ€–(10)ใ€—^10)/(๐‘ฅ โˆ’ 1) Using (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’๐‘Ž) ( ๐‘ฅ^๐‘› โˆ’ ๐‘Ž^๐‘›)/(๐‘ฅ โˆ’ ๐‘Ž) = nan โ€“ 1 Using (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’๐‘Ž) ( ๐‘ฅ^๐‘› โˆ’ ๐‘Ž^๐‘›)/(๐‘ฅ โˆ’ ๐‘Ž) = nan โ€“ 1 (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) (๐‘ฅ^15 โˆ’ ใ€–(1)ใ€—^15)/(๐‘ฅ โˆ’ 1) = 15(1)15 โ€“ 1 = 15 (1)14 = 15 (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) (๐‘ฅ^10 โˆ’ ใ€–(1)ใ€—^10)/(๐‘ฅ โˆ’ 1) = 10(1)10 โ€“ 1 = 10 (1)9 = 10 Hence , (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) (๐‘ฅ^15 โˆ’ 1^15)/(๐‘ฅ โˆ’ 1) รท (๐‘™๐‘–๐‘š)โ”ฌ(๐‘ฅโ†’1) (๐‘ฅ^10 โˆ’110)/(๐‘ฅ โˆ’ 1) = 15 รท 10 = 15/10 = 3/2 โˆด (๐’๐’Š๐’Ž)โ”ฌ(๐’™โ†’๐Ÿ) (๐’™^๐Ÿ๐Ÿ“ โˆ’ ๐Ÿ)/(๐’™^๐Ÿ๐ŸŽ โˆ’ ๐Ÿ) = ๐Ÿ‘/๐Ÿ

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