Example 3 (ii) - Chapter 12 Class 11 Limits and Derivatives

Last updated at August 4, 2026 by

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Transcript

Example 3 Evaluate: (ii) (š‘™š‘–š‘š)ā”¬(š‘„ā†’0) (āˆš(1 + š‘„) āˆ’ 1)/š‘„ (š‘™š‘–š‘š)ā”¬(š‘„ā†’0) (āˆš(1 + x )āˆ’ 1)/x Putting x = 0 = (āˆš(1 + 0) āˆ’ 1)/0 = (āˆš(1 ) āˆ’ 1)/0 = (1 āˆ’ 1)/0 = 0/0 Since it is a 0/0 form We simplify the equation Putting y = 1 + x ā‡’ y ā€“ 1 = x As x ā†’ 0 y ā†’ 1 + 0 y ā†’ 1 So, our equation becomes (š‘™š‘–š‘š)ā”¬(š‘„ā†’0) (āˆš(1 + š‘„ )āˆ’ 1)/š‘„ = (š‘™š‘–š‘š)ā”¬(š‘¦ā†’1) (āˆšš‘¦ āˆ’ 1)/(š‘¦ āˆ’ 1) = (š‘™š‘–š‘š)ā”¬(š‘„ā†’1) ( š‘¦^((āˆ’1)/2) āˆ’ 1)/(š‘¦ āˆ’ 1) = (š‘™š‘–š‘š)ā”¬(š‘„ā†’1) ( š‘¦^((āˆ’1)/2) āˆ’ 1^((āˆ’1)/2))/(š‘¦ āˆ’ 1) = 1/2 Ɨ 1^((āˆ’1)/2 āˆ’ 1) = 1/2 Ɨ 1 = šŸ/šŸ

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