Slide34.JPG

Slide35.JPG


Transcript

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

Go Ad-free
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.