Taylor series to evaluate limit

Question

Use Taylor series to evaluate lim x-> 0 [(ln√(x+1) - tan^(-1)x)] / x

Answer 1

why

as in why use Taylor Series?

as x->0 have 0/0 indeterminate form which can be resolved

with L'Hopital rule.

Derivative of numerator as x->0

is (1/2)/((1+0))-1/(1+0²)= -1/2

Derivative of denominator as x->0

is 1.

limit of expression x->0 is (-1/2)/1= -1/2

perhaps Taylor series of numerator divided by x

would result in same answer.