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

Slide31.JPG
Slide32.JPG

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

Transcript

Ex13.1, 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‘₯) ) = 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‘₯ . 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) = ((𝟏) π‘Ž + 𝑏)/(π‘Ž +(𝟏)𝑏) = (π‘Ž + 𝑏)/(π‘Ž + 𝑏) = 1

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail