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

Example 3 - Chapter 13 Class 11 Limits and Derivatives - Part 6
Example 3 - Chapter 13 Class 11 Limits and Derivatives - Part 7

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Transcript

Example 3 Evaluate: (ii) (π‘™π‘–π‘š)┬(π‘₯β†’0) (√(1 + π‘₯) βˆ’ 1)/π‘₯ (π‘™π‘–π‘š)┬(π‘₯β†’0) (√(1 + x )βˆ’ 1)/x Putting x = 0 = (√(1 + 0) βˆ’ 1)/0 = (√(1 ) βˆ’ 1)/0 = (1 βˆ’ 1)/0 = 0/0 Since it is a 0/0 form We simplify the equation Putting y = 1 + x β‡’ y – 1 = x As x β†’ 0 y β†’ 1 + 0 y β†’ 1 So, our equation becomes (π‘™π‘–π‘š)┬(π‘₯β†’0) (√(1 + π‘₯ )βˆ’ 1)/π‘₯ = (π‘™π‘–π‘š)┬(𝑦→1) (βˆšπ‘¦ βˆ’ 1)/(𝑦 βˆ’ 1) = (π‘™π‘–π‘š)┬(π‘₯β†’1) ( 𝑦^((βˆ’1)/2) βˆ’ 1)/(𝑦 βˆ’ 1) = (π‘™π‘–π‘š)┬(π‘₯β†’1) ( 𝑦^((βˆ’1)/2) βˆ’ 1^((βˆ’1)/2))/(𝑦 βˆ’ 1) = 1/2 Γ— 1^((βˆ’1)/2 βˆ’ 1) = 1/2 Γ— 1 = 𝟏/𝟐

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.