Asked by richard
How would you use series to evaluate the lim(x->0) of (x-arctanx)/x^3? I'm getting stuck, and the answer says it's 1/3 but I don't know how they got there.
Answers
Answered by
Steve
arctan x = x - x^3/3 + x^5/5 - ...
so,
x-arctan(x) = x^3/3 - x^5/5 + x^7/7 - ...
divide by x^3 and you get
1/3 - x^2/5 + x^5/7 - ...
as x->0, that -> 1/3
so,
x-arctan(x) = x^3/3 - x^5/5 + x^7/7 - ...
divide by x^3 and you get
1/3 - x^2/5 + x^5/7 - ...
as x->0, that -> 1/3
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