What is the limit of (7x^2+x-100)/(2x^2-5x) as x approaches infinity?
1 Answer
Explanation:
Notice that you're dealing with the ratio of two polynomials that have the same degree, which means that the limit when
In your case, you have
(7x^color(red)(2) + x - 100)/(2x^color(red)(2) - 5x)
The highest degree terms are
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
