How can I prove that limx→0x−sinxx3=16
without using L'Hospital or Taylor series?
thanks :)
Answer
Let L=limx→0x−sin(x)x3. We then have
L=limy→03y−sin(3y)27y3=limy→03y−3sin(y)+4sin3(y)27y3⏟sin(3y)=3sin(y)−4sin3(y)=limy→03y−3sin(y)27y3+427limy→0sin3(y)y3=327L+427
This gives us 24L = 4 \implies L = \dfrac16
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