Question #7b587

2 Answers
Nov 10, 2016

#1/3#

Explanation:

Making #x = y^3#

#(x^(1/3) +1)/(x+1) equiv (y+1)/(y^3+1)# but by the polynomial identity

#(a^3+1)/(a+1)=a^2-a+1# we have

#(x^(1/3) +1)/(x+1) equiv 1/(y^2-y+1)#

If we are interested only in real solutions, then

#lim_(x->-1)(x^(1/3) +1)/(x+1) equiv lim_(y->-1)1/(y^2-y+1) = 1/3#

Nov 10, 2016

1/3

Explanation:

The indeterminate form is 0/0.

L'Hospital's rule can be applied.

The limit is

#lim x to -1 ((x^(1/3)+1)^')/((x+1)')=lim x to -1 ((1/3)(x^(-2))^(1/3))/1=((-1)^2)^(1/3)=1/3#