## anonymous 3 years ago limit problem

1. anonymous

ok

2. anonymous

lim x approaches - infinity of 2x divided by the square root of x squared + 1

3. anonymous

$\lim_{x \rightarrow \infty} \frac{2x}{\sqrt{x^2 + 1}}$

4. anonymous

negative infinity

5. anonymous

oh

6. anonymous

In any case, how would you make the terms in a form similar to 1/x (so you can get the limits)?

7. anonymous

i'm not sure rationalize

8. anonymous

I wouldn't bother rationalizing the denominator. Waste of time. Try dividing by the least common multiple...I think that's what it's called? idk but it's x^2 so divide both numerator and denominator by x^2

9. Callisto

Divide by x instead, but this question is a little ''tricky'', I think

10. anonymous

Why divide by x?

11. Callisto

$\lim_{x \rightarrow -\infty} \frac{2x}{\sqrt{x^2 + 1}}$$=\lim_{x \rightarrow -\infty} \frac{\frac{2x}{x}}{\frac{\sqrt{x^2 + 1}}{x}}$$=-\lim_{x \rightarrow -\infty} \frac{2}{\sqrt{\frac{x^2 + 1}{x^2}}}$$=-\lim_{x \rightarrow -\infty} \frac{2}{\sqrt{1+\frac{1}{x^2}}}$$=-\frac{2}{\sqrt{1+0}}=...$I'm not quite sure since when I should put that -ve sign. But if I remember correctly, I have to put a -ve sign :|

12. anonymous

oh, that's what i meant by sqrt(x^2) so it is x

13. anonymous

I don't think you need a negative sign because it's x^2

14. Callisto

No.. I think I need it (I've got similar question wrong for n times, so I remember it :\)

15. anonymous

Oh, well the numerator is negative but the denominator is positive so it is negative