## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 4 years ago can i solve $\lim_{x \rightarrow \infty} (2-x)/\sqrt(9x^2 +1)$ without using l'hopitals rule?

• This Question is Closed
1. anonymous

sure you can do it with your eyeballs

2. anonymous

you mean algebra?

3. anonymous

numerator is like -x, denominator is like 3x and $\frac{-x}{3x}=-\frac{1}{3}$

4. anonymous

no i meant eyeballs. just by looking

5. ash2326

yeah take x^2 out of the square root $(2-x) / (x \sqrt{9+1/x^2})$ take x out from numerator $x(2/x-1) / (x \sqrt{9+1/x^2})$ $(2/x-1) / ( \sqrt{9+1/x^2})$ 1/x and 1/x^2 ---->0 as x---> $$\infty$$ so we've $-1/ \sqrt 9$ -1/3

6. anonymous

that is ok, but seems like a lot of work

7. anonymous

thanks ash for explanation, satellite your wit is endless

8. anonymous

but not helpful

9. mathmate

:)

10. anonymous

actually i was quite serious. suppose you wanted $\lim_{x\rightarrow \infty}\frac{2x^2+3x+1}{5x^2+x-2}$ wouldn't you say $\frac{2}{5}$ just like that?

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy