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What is the answer to this math question?

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

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

1 Answer
Jun 17, 2018

Answer:

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

Explanation:

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#