Ex 13.1, 20 - Evaluate: lim x->0 sin ax + bx/ ax + sin bx

Ex 13.1, 20 - Chapter 13 Class 11 Limits and Derivatives - Part 2

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Ex 12.1, 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 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.