What is the limit of f(x)=(3x^2+2) / (2x^2 -3) as x approaches -infinity and +infinity?

Question

i am so confused! and I'll get my book this week so I have no idea what to do!.help!

Answer 1

f(x)=(3x^2+2) / (2x^2 -3) as x approaches -infinity and +infinity
= (3 + 2/x^2)/(2-3/x^2).....divide top & btm by highest power of x
= (3+0)/(2+0) as x approaches -infinity and +infinity
= 3/2

=> the limit of f(x) = 3/2

Answer 2

Anytime you see a limit where the result approaches +/- infinity as the variable approaches some value, make a new fraction by taking the derivitives of the numerator and denominator. In your case the problem becomes 6x/4x. Since this too approaches infinity as x -> +/- infinity, repeat the process. Then you have 6/4, which is the answer.

Answer 3

First, divide both the top and bottom by x^2:

f = ((3x^2 + 2)/x^2) / ((2x^2 - 3)/x^2)
f = (3 + 2/x^2) / (2 - 3/x^2)

As x approaches both + and - infinite, 3/x^2 and 2/x^2 approach 0, so the answer is:

f = (3 + 0) / (2 - 0) = 3/2

Answer 4

get the leading terms:
3x^2/2x^2
because the n are the same (x^n)
you use the coefficients which are 3 and 2
so the +/- infinity limit is 3/2

Answer 5

(3x^2 + 2)/(2x^2 - 3) =
(3x^2 - 4.5 + 6.5)/(2x^2 - 3) =
= (3x^2 - 4.5)/(2x^2 - 3) + 6.5/(2x^2 - 3) =
= 1.5*(2x^2 - 3)/(2x^2 - 3) + 6.5/(2x^2 - 3) =
= 1.5 + 6.5/(2x^2 - 3)

lim as x->+-inf of (3x^2 + 2)/(2x^2 - 3) =
= lim as x->+-inf of 1.5 + 6.5/(2x^2 - 3) =
= 1.5