I understand I need to make the sum of the individual limits equal 0 – but I'm a little lost. I computed the limit of the first term to be -4/3 via L'Hospitals Rule but Wolframalpha contradicts me (http://www.wolframalpha.com/input/?i=limit+x-%3E+0+%28sin%282x%29%2F%28x%5E3%29%29).
'a' obviously should be left for last – and $b/($x^2) is some constant – so I get 0? Assuming 'b' is positive, or anything other than 0, I get that term is 0.
Thus, -4/3 + 0 + a = 0
Simple algebraic manipulating would lead me to believe a = 4/3 and then I'd plug that back in to find 'b'.
Questions
Why does Wolfram state what it does?
Is this solution correct? (I don't have the answer for the problem.)
Best Answer
We have by the Taylor series $$\sin(2x)=2x-\frac{4}{3}x^3+o(x^3)$$ so it's clear that for $b=-2$ and $a=\frac{4}{3}$ the limit is $0$.