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  1. Chapter 13 Class 11 Limits and Derivatives
  2. Serial order wise

Transcript

Ex 13.1, 20 Evaluate the Given limit: lim┬(xβ†’0) (π‘ π‘–π‘›β‘π‘Žπ‘₯ + 𝑏π‘₯)/(π‘Žπ‘₯ + 𝑠𝑖𝑛⁑𝑏π‘₯ ) a , b, a + b β‰  0 lim┬(xβ†’0) (π‘ π‘–π‘›β‘π‘Žπ‘₯ + 𝑏π‘₯)/(π‘Žπ‘₯ +γ€– 𝑠𝑖𝑛〗⁑𝑏π‘₯ ) = lim┬(xβ†’0) π‘₯(π‘ π‘–π‘›β‘π‘Žπ‘₯/π‘₯ + 𝑏)/π‘₯(π‘Ž + 𝑠𝑖𝑛⁑𝑏π‘₯/π‘₯) = lim┬(xβ†’0) ((π‘ π‘–π‘›β‘π‘Žπ‘₯/π‘₯ ) + 𝑏)/(π‘Ž + ( 𝑠𝑖𝑛⁑𝑏π‘₯/π‘₯) ) Multiply & Divide by π‘ π‘–π‘›β‘π‘Žπ‘₯/π‘₯ by ax & Multiply & Divide 𝑠𝑖𝑛⁑〖π‘₯ γ€—/𝑏 by bx = lim┬(xβ†’0) ((π‘ π‘–π‘›β‘π‘Žπ‘₯/π‘₯ . π‘Žπ‘₯/π‘Žπ‘₯ ) + 𝑏)/(π‘Ž + ( 𝑠𝑖𝑛⁑𝑏π‘₯/π‘₯ . 𝑏π‘₯/𝑏π‘₯) ) = lim┬(xβ†’0) ((π‘ π‘–π‘›β‘π‘Žπ‘₯/π‘Žπ‘₯ . π‘Žπ‘₯/π‘₯ ) + 𝑏)/(π‘Ž + ( 𝑠𝑖𝑛⁑𝑏π‘₯/𝑏π‘₯ . 𝑏π‘₯/π‘₯) ) = lim┬(xβ†’0) ((π’”π’Šπ’β‘π’‚π’™/𝒂𝒙). π‘Ž + 𝑏)/(π‘Ž + ( π’”π’Šπ’β‘π’ƒπ’™/𝒃𝒙) 𝑏) Using lim┬(xβ†’0) (sin⁑x )/x = 1 Replacing x by ax. lim┬(xβ†’0) sinβ‘π‘Žπ‘₯/ax = 1 Replacing x by bx lim┬(xβ†’0) (sin⁑bx )/bx = 1 = ((𝟏) π‘Ž + 𝑏)/(π‘Ž +(𝟏)𝑏) = (π‘Ž + 𝑏)/(π‘Ž + 𝑏) = 1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.