limit question?

Question

What values of a and b exist for this equation?


lim [(x^3-8)\(x^2-4) - (ax+b)] =0
x-> infinity

i don't know how i'm supposed to solve this.
I do know how to solve limit problems but the ax+b
and the =0 is baffling me

please show all your work
thanks!

Answer 1

You need to simplify the terms inside the brackets. You can do this my multiplying the second term by (x^2-4) so that you can get one fraction:

lim [(x^3-8)\(x^2-4) - (ax^3+bx^2-4ax-4b)/(x^2-4)] =0
x-> infinity

When we simplify this, we get:

lim [((1-a)x^3+bx^2-4ax-(8+4b))/(x^2-4)] =0
x-> infinity

Now we can consider what happens when x==> inf

On the top of the fraction, we have an x^2 term. If we divide that by an x^2 term, it leaves an x term -- so that term will go off to infinity unless (1-a) = 0.

So -- a = -1

As x==> infinity, the bx^2 term goes to b
As x==> infinity, the 4ax term goes to zero
As x==> infinity, the (8+4b) term goes to zero

Therefore, the whole thing only goes to zero if b = 0