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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.