Question

Take the limit at -infinity of:

Answer 1

\[\lim_{x \rightarrow -\infty}\frac{ x }{ x-\sqrt{x ^{2}+9 }}\]

Answer 2

My intuitive guess is -∞, but maybe L'Hopital's rule?

Answer 3

I don't believe so. I think I have to multiply by x+sqrt(x^2+9) top and bottom? Then divide each term by x^2?

Answer 4

That will help get the ↓ out of the denominator. Try it and see what you get.

Answer 5

well, I tried to just divide all terms by x^2 and I get 0/-sqrt(1) so, I guess it's 0?

Answer 6

Nope, wrong answer.

Answer 7

What do you mean, exactly, by divide all terms by x^2?

Answer 8

well, I tried to divide all terms by x^2 then plug in -infinity then the numerator would be 0 because 1/-inf is 0, the same thing would happen in the denominator with x, then x^2/x^2 is 1 and 9/x^2 is 0 when x=-infinity, but that didn't work

Answer 9

oh, no, it would be +inf not 0, right?

Answer 10

is this what you got after rationalizing the denominator?
\[\large \frac{x^2+x\sqrt{x^2+9}}{-9}\]

Answer 11

I think the answer is 2

Answer 12

yeah, but I disregarded the 9 since it's not a significant source of change.

Answer 13

so I got \[\lim_{x \rightarrow -\infty} \frac{ x }{ x-\sqrt{x^2}}\]

Answer 14

then I rationalized the denominator

Answer 15

and got \[\frac{2x ^{2} }{ x+x ^{2}}\]

Answer 16

I disregarded the x

Answer 17

and then ended up with 2

Answer 18

Does that sound about right?

Answer 19

-2!

Answer 20

Something looks fishy about that.
If you had \[\large\frac{x}{x-\sqrt{x^2}}\]
Why wouldn't you just simplify the bottom to x - x?

Answer 21

no, because you end up with x/0

Answer 22

so you rationalize and get \[\frac{ 2x ^{2} }{ x-\sqrt{x ^{2}} }\]

Answer 23

woops, no radical on the bottom

Answer 24

\[\frac{ 2x ^{2} }{ x-x ^{2} }\]

Answer 25

\[\large (x-\sqrt{x^2})(x+\sqrt{x^2}) = x-x^2?\]

Answer 26

oh no!

Answer 27

it's -1 then

Answer 28

x^2-x^2=-2x^2 so you end up with -1

Answer 29

I think it's actually 1/2.
Let me double-check that, though...

Answer 30

bear with me, \[\large\frac{-∞}{-∞ - sqrt{(-∞)^2}} = \frac{(-1)(∞)}{(-2)(∞)} = \frac{1}{2}\]
looks totally bogus, I know, but it works.

Answer 31

Remember, it's limit as x approaches -∞, it's not actually ∞/∞, pretend it's just some number over that same number.

Answer 32

oh, I understand now!

Answer 33

I get it!