I'm unsure as to how to evaluate:
limx→0sinx−x+x36x3
The limx→0 of both the numerator and denominator equal 0. Taking the derivative of both ends of the fraction we get:
limx→0x2+2cosx−26x2
But I don't know how to evaluate this?
Many thanks for any help.
Answer
You can use l'Hospital as many times as needed as long as the indeterminate forms conditions are fulfilled. In this case, using Taylor series can be helpful, too:
sinx=x−x36+x5120−…=x−x36+O(x5)
⟹sinx−x+x36x3=O(x5)x3=O(x2)→x→00
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