# Limits Help

• Nov 18th 2012, 01:40 PM
NettieL
Limits Help
Hi there,
just pondering this equation: lim (3x^3(-6x^2)+5x+35)/((x^4)-5x^3-27) when x->∞

what is the best way to figure this out? I tried to divide everything by the highest factor I would find in the denominator, like my textbook said, but that just got messy and confusing. Any ideas?
• Nov 18th 2012, 03:14 PM
MarkFL
Re: Limits Help
Is the limit actually:

$\displaystyle \lim_{x\to\infty}\frac{3x^3-6x^2+5x+35}{x^4-5x^3-27}$ ?

If so, you might just observe that the limit is zero since the degree in the denominator is greater than that of the numerator.
• Nov 18th 2012, 05:59 PM
NettieL
Re: Limits Help
oh, ok then, that means that I did get it right (I came out with 0/1), but what confused me was that wolfram indicated it was -infinity. Glad I seem to have done the right thing then!!
• Nov 18th 2012, 06:12 PM
Prove It
Re: Limits Help
I don't see how you could have gotten \displaystyle \displaystyle \begin{align*} \frac{0}{1} \end{align*}. The reason the limit is \displaystyle \displaystyle \begin{align*} 0 \end{align*} is because the denominator goes to \displaystyle \displaystyle \begin{align*} \infty \end{align*} while the numerator does not. Dividing a number by a much larger number gives you something very close to 0.
• Nov 18th 2012, 06:16 PM
MarkFL
Re: Limits Help
If you typed it into the text box at wolfram as you did in your first post, then you would get -infinity. Do you see why?
• Nov 18th 2012, 06:19 PM
NettieL
Re: Limits Help
ahh, it was all those unnecessary brackets I put in... silly me :)
• Nov 18th 2012, 06:27 PM
NettieL
Re: Limits Help
in that particular wolfram working, it divides each term by a different power... why is this? I thought we would only divide by x^4??Attachment 25801
• Nov 18th 2012, 06:28 PM
Prove It
Re: Limits Help
They DID divide top and bottom by \displaystyle \displaystyle \begin{align*} x^4 \end{align*}...
• Nov 18th 2012, 06:42 PM
NettieL
Re: Limits Help
so what about where they divided by x^3, x^2?? Sorry i know I sound like a complete idiot, but these things take time to sink in :)
• Nov 18th 2012, 06:50 PM
Prove It
Re: Limits Help
\displaystyle \displaystyle \begin{align*} \frac{3x^3 - 6x^2 + 5x + 35}{x^4 - 5x^3 - 27} &= \frac{\frac{3x^3 - 6x^2 + 5x + 35}{x^4}}{\frac{x^4 - 5x^3 - 27}{x^4}} \\ &= \frac{\frac{3x^3}{x^4} - \frac{6x^2}{x^4} + \frac{5x}{x^4} + \frac{35}{x^4}}{\frac{x^4}{x^4} - \frac{5x^3}{x^4} - \frac{27}{x^4}} \\ &= \frac{\frac{3}{x} - \frac{6}{x^2} + \frac{5}{x^3} + \frac{35}{x^4}}{1 - \frac{5}{x} - \frac{27}{x^4}} \end{align*}
• Nov 18th 2012, 09:24 PM
NettieL
Re: Limits Help
Oh, now I get it!! Thank you so much! :)
• Nov 19th 2012, 01:57 PM