I need to find the limit of f(x)=(x^2+6x+8)/(x+3)^2 as x approaches positive and negative infinity
I know how to solve the limit as it goes to positive infinity (I got 1) but I don't know how to solve it when it goes to negative infinity (I know the answer is 1 though). Please help
Either way will work but notice that neither Dinkydoe's nor earboth's solutions involved "dividing everything by the highest power of x"! They divided to get a polynomial plus a fraction with "1" in the numerator.
If you write , by multiplying out the square in the denominator. Now divide by numerator and denominator by the highest power of x, , to get [tex]\frac{1+ 6/x+ 8/x^2}{1+ 6/x+ 9/x^2}. Now, as x goes to or , all of those fractions go to 0 while the "1" does not depend on x so remains 1: the limit is whether x goes to infinity or negative infinity.
You could in order to determine the limit as x goes to , divide by -x to the highest power and then look as x going to but here, since that highest power is even, it doesn't matter.
When determining the end-behavior of the quotient of polynomials, I think the easiest way to understand it (and remember it) is to take the highest power term from the numerator and denominator and divide them.
So, for the problem you provided:
As another example:
And another:
Just make sure you only apply this method when dealing with end-behavior.