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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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Example 3 Evaluate: (i) (π‘™π‘–π‘š)┬(π‘₯β†’1) (π‘₯ 15 βˆ’ 1)/(π‘₯10 βˆ’ 1) (π‘™π‘–π‘š)┬(π‘₯β†’1) (π‘₯ 15 βˆ’ 1)/(π‘₯10 βˆ’ 1) = (γ€–(1)γ€—^15 βˆ’ 1)/(γ€–(1)γ€—^10 βˆ’ 1) = (1 βˆ’ 1)/(1 βˆ’ 1) = 0/0 Since it is form 0/0, We can solve by using theorem (π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) (π‘₯^𝑛 βˆ’ π‘Ž^𝑛)/(π‘₯ βˆ’ π‘Ž) = na n – 1 Hence, (π‘™π‘–π‘š)┬(π‘₯β†’1) (π‘₯^15 βˆ’ 1)/(π‘₯^10 βˆ’ 1) = (π‘™π‘–π‘š)┬(π‘₯β†’1) π‘₯^15 – 1 Γ·lim┬(xβ†’1) x10 – 1 = (π‘™π‘–π‘š)┬(π‘₯β†’1) π‘₯^15 – γ€–(1)γ€—^15 Γ· lim┬(xβ†’1) x10 – (1)10 Multiplying and dividing by x – 1 = (π‘™π‘–π‘š)┬(π‘₯β†’1) (π‘₯^15 βˆ’ 1^15)/(π‘₯ βˆ’ 1) Γ· (π‘™π‘–π‘š)┬(𝑧→1) (π‘₯^10 βˆ’ γ€–(10)γ€—^10)/(π‘₯ βˆ’ 1) Using (π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) ( π‘₯^𝑛 βˆ’ π‘Ž^𝑛)/(π‘₯ βˆ’ π‘Ž) = nan – 1 Using (π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) ( π‘₯^𝑛 βˆ’ π‘Ž^𝑛)/(π‘₯ βˆ’ π‘Ž) = nan – 1 (π‘™π‘–π‘š)┬(π‘₯β†’1) (π‘₯^15 βˆ’ γ€–(1)γ€—^15)/(π‘₯ βˆ’ 1) = 15(1)15 – 1 = 15 (1)14 = 15 (π‘™π‘–π‘š)┬(π‘₯β†’1) (π‘₯^10 βˆ’ γ€–(1)γ€—^10)/(π‘₯ βˆ’ 1) = 10(1)10 – 1 = 10 (1)9 = 10 Hence , (π‘™π‘–π‘š)┬(π‘₯β†’1) (π‘₯^15 βˆ’ 1^15)/(π‘₯ βˆ’ 1) Γ· (π‘™π‘–π‘š)┬(π‘₯β†’1) (π‘₯^10 βˆ’110)/(π‘₯ βˆ’ 1) = 15 Γ· 10 = 15/10 = 3/2 ∴ (π’π’Šπ’Ž)┬(π’™β†’πŸ) (𝒙^πŸπŸ“ βˆ’ 𝟏)/(𝒙^𝟏𝟎 βˆ’ 𝟏) = πŸ‘/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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.