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Question

Use Taylor's power series for $tan^-1x$ to evaluate:

$lim_(x->0)(tan^-1x-x)/x^3$

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Answer 1 · 2018-06-17T15:23:05

$lim_(x->0)(arctanx-x)/x^3=-1/3$

We want to find $lim_(x->0)(tan^-1x-x)/x^3=0$ using Taylor series

First, let $tan^-1-= arctan$

Now,

$arctanx=x-x^3/3+x^5/5-x^7/7$

so

$lim_(x->0)(arctanx-x)/x^3$

$=lim_(x->0)(x-x^3/3+x^5/5-...-x)/x^3$

$=lim_(x->0)(-1/3+x^2/5-x^4/7...)$

$=-1/3$