Question

What is the limit of `(7x^2+x-100)/(2x^2-5x)` as x approaches infinity?

Answer 1

`lim_(x->oo)f(x) = 7/2`

Explanation:

Notice that you're dealing with the ratio of two polynomials that have the same degree, which means that the limit when `x->oo` will be equal to the ratio of the coefficients that belong to the terms that have the highest degree.

In your case, you have

`(7x^color(red)(2) + x - 100)/(2x^color(red)(2) - 5x)`

The highest degree terms are `7x^2` and `2x^2`, which means that the limit will be

`lim_(x->oo)(color(blue)(7)x^2 + x - 100)/(color(blue)(2)x^2 - 5x) = color(blue)(7)/color(blue)(2)`

Alternatively, you can prove this by doing some algebraic manipulation of the fraction's numerator and denominator. More specifically, you can write

`7x^2 + x - 100 = x^2 * (7 + 1/x - 100/x^2)`

and

`2x^2 - 5x = x^2 * (2 - 5/x)`

The limit now becomes

`lim_(x->oo)(color(red)(cancel(color(black)(x^2))) * (7 + 1/x - 100/x^2))/(color(red)(cancel(color(black)(x^2))) * (2 - 5/x)) = lim_(x->oo)(7 + 1/x - 100/x^2)/(2 - 5/x)`

This is equal to

`lim_(x->oo)(7 + 1/x - 100/x^2)/(2 - 5/x) = (7 + 0 + 0)/(2 + 0) = 7/2`