Lim x->0 [cosx+a³sin(b^6x)]^1/x = e^512 .find the value of ab²
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1
Answer:
Step-by-step explanation:
Answer:
value of ab² = 8³
Step-by-step explanation:
Lim (1+casx+a3 sin (b6x) - 1)0
x = 0
casx +a3 sin (b6x)-1
x
Lim 1-x2 + b6 a3 sin (b6x) - 1/x
2 xb6
x
l b6 a3 = l512
b6 a3 = 512
(AB2)3 = 8³
Answered by
0
Answer:
Value of ab² is 8
In the last step I have taken log on both sides
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