Ex 12.1, 20 - Chapter 12 Class 11 Limits and Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Ex 12.1
Ex 12.1, 1
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Transcript
Ex 12.1, 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
