# Thread: limit to infinity

1. ## limit to infinity

How do I do this?

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

Any help?

2. it goes to $\displaystyle \infty$
the numerator is bigger than denominator

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

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

Originally Posted by javax
it goes to $\displaystyle \infty$
the numerator is bigger than denominator
Not exactly. $\displaystyle \sqrt{x^3} = x^{\frac{3}{2}} \ < \ x^2$

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

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

Not exactly. $\displaystyle \sqrt{x^3} = x^{\frac{3}{2}} \ < \ x^2$
opps, I didn't notice the root at all! Sorry

5. I tried using the sandwich rule but my final answer doesn't make much sense:

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?

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

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 $\displaystyle \lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} h(x) = L$ then $\displaystyle \lim_{x \rightarrow a} g(x) = L$.

7. Oh right. It just didn't look right when I got to the end.

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