What is the limit of (3x^2-8x+1)/(2-7x^2) as x goes to infinity?

1 Answer
Sep 26, 2015

lim_(xrarroo)(3x^2-8x+1)/(2-7x^2) = -3/7

Explanation:

An attempt to use the quotient rule for limits at infinity gets us the indeterminate form oo/oo.

However, (3x^2-8x+1)/(2-7x^2) can be rewritten for x !=0.
(when we ask about limits at infinity, we can safely ignore what happens when x=0)

The rewritten fraction will have a denominator with a finite limit.

There are at least three ways to describe this rewriting.

1) Divide the numerator and denominator by the greatest power of x in the denominator. In this case divide both by x^2

2) Multiply numerator and denominator be 1 over the greatest power of x in the denominator. In this case multiply both by 1/x^2

3) Factor out the greatest power of x in the denominator from both the numerator and the denominator and then reduce the fraction.

I learned and am still most comfortable with description 3).

(3x^2-8x+1)/(2-7x^2) = (x^2(3-8/x+1/x^2))/(x^2(2/x^2-7))

= (cancel(x^2)(3-8/x+1/x^2))/(cancel(x^2)(2/x^2-7))

lim_(xrarroo) (3x^2-8x+1)/(2-7x^2) = lim_(xrarroo) (3-8/x+1/x^2)/(2/x^2-7)

= (3-0+0)/(0-7) = -3/7

Remember the key idea is to rewrite so that the denominator has a finite limit.

Example 1
lim_(xrarroo)(2x^5+4x^2-9)/(3x^5+6x+1) = lim_(xrarroo)(cancel(x^5)(2+4/x^3-9/x^5))/(cancel(x^5)(3+6/x^4+1/x^5)) = 2/3

Example 2
lim_(xrarroo)(x^3+2x^2-7x+6)/(x^2+x+9) = lim_(xrarroo)(cancel(x^2)(x+2-7/x+6/x^2))/(cancel(x^2)(1+1/x+9/x^2)) " " which has the form (oo+2)/1, so we write

lim_(xrarroo)(x^3+2x^2-7x+6)/(x^2+x+9) = oo

Example 3

lim_(xrarroo)(4x^2+2x-1)/(2x^3-7x^2+8) = " what?"

find the greatest power of x in the denominator.

factor it pout of the numerator and the denominator, and reduce

so the denominator no longer has an infinite limit

the limit of the new denominator is 2

what is the limit of the new denominator?

If you said 0, good

Now what is 0/2?

If you said 0 again, you're correct. So the limit is 0
Take a break and have a cookie! Good job.