# `lim_(x -> -oo) (4x^2 + 5)/(x^2 + 3)` Find the limit.

### Textbook Question

Chapter 3, 3.5 - Problem 22 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to evaluate the limit, hence, you need to replace `oo` for x in equation:

`lim_(x->-oo) (4x^2 + 5)/(x^2 + 3) = (4oo + 5)/(oo + 3) = (oo)/oo`

Since the result is indeterminate, you need to force `x^2` factors at numerator and denominator:

`lim_(x->-oo) (x^2(4 + 5/(x^2)))/(x^2(1 + 3/(x^2)))`

Since l`im_(x->oo) 5/(x^2) = 0 ` and` lim_(x->oo) 3/(x^2) = 0` , yields:

`lim_(x->-oo) (x^(2-2))*(4/1)= 4*lim_(x->oo) (x^0) = 4*1 = 4`

Hence, evaluating the given limit yields `lim_(x->-oo) (4x^2 + 5)/(x^2 + 3) = 4.`

scisser | (Level 3) Honors

Posted on

Plugging in `oo ` , you get

`lim_(x->oo)(4(oo)^2+5)/((oo)^2+3)=oo/oo `

Use L^Hopital's Rule, taking the derivative of the numerator and denominator independently.

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

Use LH's Rule again.

`lim_(x->oo)8/2=4 `

Therefore, the `lim_(x->oo)(4x^2+5)/(x^2+3)=4`