Subscribe to our Youtube Channel - https://you.tube/teachoo
Ex 13.1, 20 - Chapter 13 Class 11 Limits and Derivatives
Last updated at May 4, 2020 by Teachoo
Transcript
Ex 13.1, 20 Evaluate the Given limit: limβ¬(xβ0) (π ππβ‘ππ₯ + ππ₯)/(ππ₯ + π ππβ‘ππ₯ ) a , b, a + b β 0 limβ¬(xβ0) (π ππβ‘ππ₯ + ππ₯)/(ππ₯ +γ π ππγβ‘ππ₯ ) = limβ¬(xβ0) π₯(π ππβ‘ππ₯/π₯ + π)/π₯(π + π ππβ‘ππ₯/π₯) = limβ¬(xβ0) ((π ππβ‘ππ₯/π₯ ) + π)/(π + ( π ππβ‘ππ₯/π₯) ) Multiply & Divide by π ππβ‘ππ₯/π₯ by ax & Multiply & Divide π ππβ‘γπ₯ γ/π by bx = limβ¬(xβ0) ((π ππβ‘ππ₯/π₯ . ππ₯/ππ₯ ) + π)/(π + ( π ππβ‘ππ₯/π₯ . ππ₯/ππ₯) ) = limβ¬(xβ0) ((π ππβ‘ππ₯/ππ₯ . ππ₯/π₯ ) + π)/(π + ( π ππβ‘ππ₯/ππ₯ . ππ₯/π₯) ) = limβ¬(xβ0) ((πππβ‘ππ/ππ). π + π)/(π + ( πππβ‘ππ/ππ) π) Using limβ¬(xβ0) (sinβ‘x )/x = 1 Replacing x by ax. limβ¬(xβ0) sinβ‘ππ₯/ax = 1 Replacing x by bx limβ¬(xβ0) (sinβ‘bx )/bx = 1 = ((π) π + π)/(π +(π)π) = (π + π)/(π + π) = 1
Ex 13.1
Ex 13.1, 1
Ex 13.1, 2
Ex 13.1, 3
Ex 13.1, 4
Ex 13.1, 5
Ex 13.1, 6 Important
Ex 13.1, 7
Ex 13.1, 8 Important
Ex 13.1, 9
Ex 13.1,10 Important
Ex 13.1, 11
Ex 13.1, 12
Ex 13.1, 13
Ex 13.1, 14 Important
Ex 13.1, 15 Important
Ex 13.1, 16
Ex 13.1, 17
Ex 13.1, 18
Ex 13.1, 19
Ex 13.1, 20 You are here
Ex 13.1, 21 Important
Ex 13.1, 22 Important
Ex 13.1, 23
Ex 13.1, 24
Ex 13.1, 25 Important
Ex 13.1, 26
Ex 13.1, 27
Ex 13.1, 28 Important
Ex 13.1, 29 Important
Ex 13.1, 30 Important
Ex 13.1, 31 Important
Ex 13.1, 32 Important
About the Author
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.