Find the limit. (If you need to use or –, enter INFINITY or –INFINITY, respectively.)?lim x→− Infinity (−2x + 8) / √(x2 + x)

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to evaluate the following limit such that:

`lim_(x->-oo) (-2x + 8) /(sqrt(x^2+x)) = oo/oo`

You may force factor x to numerator such that:

`lim_(x->-oo) x(-2 + 8/x)/(sqrt(x^2+x))`

You may force factor `x^2`  to denominator such that:

`lim_(x->-oo) x(-2 + 8/x)/(sqrt(x^2(1+x/x^2)))`

You should remember that `sqrt(x^2) = |x|`  and since `x->-oo ` `=> |x| = -x`  such that:

`lim_(x->-oo) x(-2 + 8/x)/(-xsqrt(1+1/x))`

Reducing x yields:

`lim_(x->-oo) (-2 + 8/x)/(-sqrt(1+1/x)) = (-2 + lim_(x->-oo) (8/x))/(-sqrt(1 + lim_(x->-oo) (1/x)))`

`lim_(x->-oo) (-2 + 8/x)/(-sqrt(1+1/x)) = (-2 + 0)/(-sqrt(1+0))`

`lim_(x->-oo) (-2 + 8/x)/(-sqrt(1+1/x)) = (-2)/(-1) = 2`

Hence, evaluating the given limit yields `lim_(x->-oo) (-2x + 8) /(sqrt(x^2+x)) = 2.`

We’ve answered 318,983 questions. We can answer yours, too.

Ask a question