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.

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

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