# limit to infinity

• Sep 21st 2008, 07:05 PM
kitle545
limit to infinity
How do I do this?

lim (square root(8x^3 + 5x + 10)) / (x^2)
x->infinity

Any help?
• Sep 21st 2008, 07:13 PM
javax
it goes to $\infty$
the numerator is bigger than denominator
• Sep 21st 2008, 07:14 PM
o_O
$\lim_{x \to \infty} \frac{\sqrt{8x^3 + 5x + 10}}{x^2} \cdot \frac{\displaystyle {\color{red}\frac{1}{x^2}}}{\ \displaystyle{\color{red}\frac{1}{x^2}} \ } \ = \ \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{x^2}}{\displaystyle \frac{\ x^2 \ }{x^2}}$

$= \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{\sqrt{x^4}}}{1} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3 + 5x + 10}{x^4}} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3}{x^4} + \frac{5x}{x^4} + \frac{10}{x^4}} \ = \ \hdots$

Quote:

Originally Posted by javax
it goes to $\infty$
the numerator is bigger than denominator

Not exactly. $\sqrt{x^3} = x^{\frac{3}{2}} \ < \ x^2$
• Sep 21st 2008, 08:25 PM
javax
Quote:

Originally Posted by o_O
$\lim_{x \to \infty} \frac{\sqrt{8x^3 + 5x + 10}}{x^2} \cdot \frac{\displaystyle {\color{red}\frac{1}{x^2}}}{\ \displaystyle{\color{red}\frac{1}{x^2}} \ } \ = \ \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{x^2}}{\displaystyle \frac{\ x^2 \ }{x^2}}$

$= \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{\sqrt{x^4}}}{1} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3 + 5x + 10}{x^4}} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3}{x^4} + \frac{5x}{x^4} + \frac{10}{x^4}} \ = \ \hdots$

Not exactly. $\sqrt{x^3} = x^{\frac{3}{2}} \ < \ x^2$

opps, I didn't notice the root at all! Sorry
• Sep 22nd 2008, 05:03 AM
Showcase_22
I tried using the sandwich rule but my final answer doesn't make much sense:

http://i116.photobucket.com/albums/o...sproblem71.jpg

The final bit doesn't make sense since I get that it convereges to a number between 0 and 0 that isn't 0. :s

any helpers?
• Sep 22nd 2008, 06:18 AM
mr fantastic
Quote:

Originally Posted by Showcase_22
I tried using the sandwich rule but my final answer doesn't make much sense:

http://i116.photobucket.com/albums/o...sproblem71.jpg

The final bit doesn't make sense since I get that it convereges to a number between 0 and 0 that isn't 0. :s

any helpers?

The sandwich theorem works fine.

If f(x) < g(x) < h(x) and $\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} h(x) = L$ then $\lim_{x \rightarrow a} g(x) = L$.
• Sep 23rd 2008, 04:05 AM
Showcase_22
Oh right. It just didn't look right when I got to the end.(Worried)

I think that's the first time i've ever used the sandwich rule. It's nice to know I was doing it correctly!(Giggle)