What is the answer to this math question?

Question

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

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

1012 views around the world

You can reuse this answer
Creative Commons License

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#