How do you evaluate the limit #(x^3+2)/(x+1)# as x approaches #oo#?

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Answer 1 · 2016-12-04T15:31:54

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

You can separate the numerator in two parts, one of which can be divided by (x+1). For instance add and subtract #1# and you get:

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

As #bar x = -1# is a root of #x^3+1#, then #(x+1)# must be a factor, and in fact:

#(x^3 +1) = (x^2-x+1)(x+1)#

so that:

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

Now go for the limit: the first addendum tends to infinity, the second addendum tends to zero.

#lim_(x->oo) (x^3+2)/(x+1) = lim_(x->oo)[(x^2-x+1) +1/(x+1)] =+oo#