Limits at Infinity

**derfliw** · October 10th 2009, 07:30 AM

Say I have a limit of a fraction where x->infinity, which should I divide the equation with? The numerator's highest degree or the denominator's highest degree?

**skeeter** · October 10th 2009, 09:13 AM

Originally Posted by derfliw

Say I have a limit of a fraction where x->infinity, which should I divide the equation with? The numerator's highest degree or the denominator's highest degree?

divide by the highest degree when numerator and denominator have the same degree. you shouldn't have to divide if one is larger than the other.

**derfliw** · October 10th 2009, 10:14 AM

What you mean is, it's optional to not divide it right?

What if you choose to divide it?

**skeeter** · October 10th 2009, 10:24 AM

Originally Posted by derfliw

What you mean is, it's optional to not divide it right?

What if you choose to divide it?

optional?

that's not what I said. once again, with a little more clarity ...

1. if the degree of the numerator = degree of the denominator, then divide every term by the highest degree variable.

2. if degree of the numerator > degree of the denominator, then the limit does not exist ... the rational function changes w/o bound.

3. if the degree of the numerator < degree of the denominator, then the limit is 0.

I believe that covers all three possibilities.

**derfliw** · October 10th 2009, 10:37 AM

what if it's like this:

lim x->infinite (x^3+x^2-1)/(x^2+1)

how'd you solve that?

**skeeter** · October 10th 2009, 11:01 AM

Originally Posted by derfliw

what if it's like this:

lim x->infinite (x^3+x^2-1)/(x^2+1)

how'd you solve that?

the degree of the numerator (3) is greater than the degree of the denominator (2) ...

now go back and read what I last posted.

**derfliw** · October 10th 2009, 11:07 AM

It's just that, I have to show that it's equal to infinity.
So err, you don't need to divide all terms by x^2?
If thats so, Thanks for the help.

**skeeter** · October 10th 2009, 11:23 AM

Originally Posted by derfliw

It's just that, I have to show that it's equal to infinity.
So err, you don't need to divide all terms by x^2?
If thats so, Thanks for the help.

divide every term by $x^3$ ...

$\lim_{x \to \infty} \frac{1 + \frac{1}{x} - \frac{1}{x^3}}{\frac{1}{x} + \frac{1}{x^3}}<br />$

as x gets very large, the numerator approaches 1 and the denominator goes to 0 ...

in essence, it is the fraction $\frac{1}{a \, very \, small \, number}$

the value of the rational function increases w/o bound.

**derfliw** · October 10th 2009, 11:28 AM

Uh, so it's not the denominator's highest degree, but instead the whole equation's highest degree. Okay, I got it now, that's what i've been meaning to ask.

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