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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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Transcript

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 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.