Question

Evaluate the limit at -infinity. x/ (x-sqrt(x^2+7))

Answer 1

i did it and yet couldnt find out

Answer 2

\[\frac{x}{x -\sqrt{x^2- 7 }}\]
\[ \frac{1}{1 - \sqrt{1 - \frac{7}{x^2}}}\]No...maybe some other approach..is the answer undefined?

Answer 3

james i tried it already.. couldnt get the answer

it should be 1/2

Answer 4

I'm trying and I can't get it either!

Answer 5

im so desperate .. its not homework or something
i just want to do it!

Answer 6

yes it comes out as undefined either way, right?

Answer 7

or the book is wrong...

Answer 8

i put it in "limit calculator" and it gives me 1/2
also when i plug really large numbers it gives me 1/2

Answer 9

really large negative* numbers

Answer 10

the limit is 1/2
i can show you

Answer 11

ill be happy

Answer 12

yes

Answer 13

Okay, hmm if I think about the equation I do get \(\frac{1}{2}\)

Answer 14

My TI-89 returns 1/2

Answer 15

\[\sqrt{x^2+7} \to x \]Right?

Answer 16

\[\lim_{x \rightarrow -\infty}\frac{x}{x-\sqrt{x^2+7}} \]

since x<0=>|x|=-x

remember \[|x|=\sqrt{x^2}\]
so anyways top and bottom by \[\sqrt{x^2}\]
\[\lim_{x \rightarrow -\infty} \frac{\frac{x}{\sqrt{x^2}}}{\frac{x}{\sqrt{x^2}}-\sqrt{\frac{x^2+7}{x^2}}}\]
\[\lim_{x \rightarrow -\infty}\frac{\frac{x}{-x}}{\frac{x}{-x}-\sqrt{1+\frac{7}{x^2}}}\]

Answer 17

And also keep in mind we have negative large numbers here ...

Answer 18

\[\frac{-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}\]

Answer 19

So it's kinds like \[\frac{-x}{-x - x}\to \frac{1}{2}\]

Answer 20

very nice .. thank you all :)

Answer 21

I used L'Hospital's Rule and got\[\lim_{x \rightarrow -\infty}1/[1-(1/\sqrt{1+7/x^2})\] which if you can figure an excuse to take the negative root in the denominator yields 1/2. That's just ad hoc though.

Answer 22

you don't have to use l'hospital's rule
but whatever floats your boat

Answer 23

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