Limit as x approaches positive & negative infinity

**rawkstar** · January 18th 2010, 04:50 AM

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

**Dinkydoe** · January 18th 2010, 04:58 AM

$\lim_{x\to-\infty}\frac{x^2+6x+8}{(x+3)^2} = \lim_{x\to-\infty}\frac{(x+3)^2-1}{(x+3)^2} = 1- \lim_{x\to-\infty}\frac{-1}{(x+3)^2} = 1$

**earboth** · January 18th 2010, 04:59 AM

Originally Posted by rawkstar

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

Do first a little bit algebra warm up before you start to determine the limit:

$f(x)=\dfrac{x^2+6x+8}{(x+3)^2} = \dfrac{x^2+6x+9-1}{(x+3)^2} \dfrac{(x+3)^2-1}{(x+3)^2} = 1-\dfrac1{(x+3)^2}$

The denominator of the fraction is positive for all $x \in \mathbb{R} \setminus\{-3\}$ . The fraction approaches zero for all $x \in \mathbb{R} \setminus\{-3\}$ . Thus the limit is always 1.

EDIT: Helaas, weer te laat!

**rawkstar** · January 18th 2010, 05:30 AM

this is really helpful but not really what im looking for
let me rephrase my question
I want to know when you divide everything by the highest power of x in the denominator, do you divide by -x or +x when x is approaching -inf

**HallsofIvy** · January 18th 2010, 05:44 AM

Originally Posted by rawkstar

this is really helpful but not really what im looking for
let me rephrase my question
I want to know when you divide everything by the highest power of x in the denominator, do you divide by -x or +x when x is approaching -inf

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 $\frac{x^2+ 6x+ 8}{(x+3)^2}= \frac{x^2+ 6x+ 8}{x^2+ 6x+ 9}$ , by multiplying out the square in the denominator. Now divide by numerator and denominator by the highest power of x, $x^2$ , to get [tex]\frac{1+ 6/x+ 8/x^2}{1+ 6/x+ 9/x^2}. Now, as x goes to $+\infty$ or $-\infty$ , all of those fractions go to 0 while the "1" does not depend on x so remains 1: the limit is $\frac{1}{1}= 1$ whether x goes to infinity or negative infinity.

You could in order to determine the limit as x goes to $-\infty$ , divide by -x to the highest power and then look as x going to $+\infty$ but here, since that highest power is even, it doesn't matter.

**drumist** · January 18th 2010, 05:54 AM

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:

$ \lim_{x\to-\infty} \frac{x^2+6x+8}{(x+3)^2} =\lim_{x\to-\infty} \frac{x^2+6x+8}{x^2+6x+9} =\lim_{x\to-\infty} \frac{x^2}{x^2} =\lim_{x\to-\infty} 1 =1 $

As another example:

$ \lim_{x\to-\infty} \frac{-2x^3+2x+1}{5x^2+3} =\lim_{x\to-\infty} \frac{-2x^3}{5x^2} =\lim_{x\to-\infty} -\frac{2}{5} x =+\infty $

And another:

$ \lim_{x\to\infty} \frac{x^2+6x+1}{-2x^4+1} =\lim_{x\to\infty} \frac{x^2}{-2x^4} =\lim_{x\to\infty} -\frac{1}{2x^2} =0 $

Just make sure you only apply this method when dealing with end-behavior.

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