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Evaluating a limit using Taylor series?
Use Taylor series to evaluate the limit:
lim(x->0) (x-arctanx)/(x^3)
I keep getting infinity, but I don't think that's correct. Please explain how to find the limit! :)
1 Answer
- kbLv 78 years agoFavorite Answer
Series for arctan x:
Since 1/(1 - t) = 1 + t + t^2 + ..., letting t = -x^2 yields
1/(1 + x^2) = 1 - x^2 + x^4 - ... .
Integrate both sides from 0 to x:
arctan x = x - x^3/3 + x^5/5 - ...
-----------------------------
Therefore,
lim(x→0) [x - arctan(x)] / x^3
= lim(x→0) [x - (x - x^3/3 + x^5/5 - ...)] / x^3
= lim(x→0) (x^3/3 - x^5/5 + ...) / x^3
= lim(x→0) x^3 (1/3 - x^2/5 + ...) / x^3
= lim(x→0) (1/3 - x^2/5 + ...)
= 1/3.
I hope this helps!