Question

lim (x - sqrt(x^2 + 7)) as x goes to negative infinity. Compute the limit.

I just get 0/0 when I go through the process of dividing by the highest power of x in the denominator and accounted for the negative in negative infinity. Is that correct?

Answer 1

multiply both top and bottom by (x + sqrt(x^2+7))

Answer 2

Would you like to see the work that i've done? I've already multiplied by the conjugate.

Answer 3

you should have had -7/(x + sqrt(x^2 + 7) yes?

Answer 4

yes, the next step being dividing the top and bottom terms by "x" correct?

Answer 5

actually no, you would not accomplish anything doing so

Answer 6

\(\bf x-\sqrt{x^2}\implies x-x\implies 0\\ \quad \\
x^2+7>x^2\qquad thus\qquad x-\sqrt{x^2+7}\implies \textit{a negative value}\)

the bigger "x" becomes, the bigger that negative value I'd think

Answer 7

wait, nvm

Answer 8

http://imgur.com/tq6vc3n this is the work i've done after multiplying by the conjugate. i've followed all the rules for infinite limits I believe. Not sure why I'm still ending up getting an indeterminate form. Perhaps an error in the question?

Answer 9

the problem is that sqrt(x^2) isn't x. This is only true when x > 0

Answer 10

lim (x - sqrt(x^2 + 7)) as x -> -infty
what is the limit as \(x\to-\infty\) of x ?
what is the limit as \(x\to-\infty\) of sqrt(x^2+7) ?

Answer 11

Negative infinity and infinity respectively.

Answer 12

nice, so that gives the limit as\[-\infty-(\infty)=-\infty\]there is no aspect of the function that does not tend to -infty, hence that's the limit

Answer 13

Would it still be correct in saying that the limit does not exist since infinity is well, you know, infinite?

Answer 14

yes, when the limit is infinity you can always say it doesn't exist, because infinity is not a number anyway

Answer 15

Thank you :]

Answer 16

welcome!

Answer 17

the limit is -infinity but how would you show that?

Answer 18

find the limit of x, then find the limit of -sqrt(x^2+7), then sum the two

Answer 19

if you got \(-\infty+\infty\), then the limit could be anything, but since *both* parts tend to \(-\infty\), so does the whole thing

Answer 20

but the whole thing is under the denominator, wouldn't it be 0? -7/(a large number) is closed to 0. But the thing is the limit as x->-inf of 7/(x + sqrt(x^2+7)) is -inf

Answer 21

I didn't do that conjugate thing, so mine is just as it was originally stated, everything in the numerator

Answer 22

if you do the conjugate thing it makes it all more complicated, because the you *do* get \[-\infty+\infty\]in the denominator, and that can take on literally *any* value. So in this case, multiplying by the conjugate actually gets us farther from the answer, not closer...

Answer 23

ahh.. I see