I need help solving this one guys:
$\displaystyle \lim_{x\to -\infty}\frac{2x}{3\sqrt{x^2+1}}$
How can I show that the limit exists and prove it?
ok so now my strategy is to use the squeeze theorem to show that the function $\displaystyle \frac{x}{|x|\sqrt{1+1/(x^2)}}$ goes to -1 as x goes to $\displaystyle -\infty$.
As you suggested, x/|x| goes to -1. My problem now is finding something smaller than $\displaystyle \frac{x}{|x|\sqrt{1+1/(x^2)}}$ which also goes to -1 as x goes to $\displaystyle -\infty$
Is this even the right strategy to take? Im a little lost.
I'm sure it's sufficient to note that $\displaystyle \frac{1}{x^2} \rightarrow 0$ in the limit of $\displaystyle x \rightarrow - \infty$ and use a couple of basic limit theorems (it appears that you have more or less done that in saying that
.$\displaystyle \frac{x}{|x|\sqrt{1+1/(x^2)}}$ goes to -1 as x goes to $\displaystyle -\infty$.
well Im trying to use the squeeze theorem, but I cant find a function smaller than x/(|x|sqrt(1+1/x^2))) which goes to -1 as x goes to -inf.
So I havent actually shown that x/(|x|sqrt(1+1/x^2))) goes to -1 yet!
I dont quite see how showing 1/x^2 going to 0 would be an effective approach, could you elaborate? Thanks for thehelp