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

1 Answer
Stefan V.
Aug 11, 2015

#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#

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