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May 30th, 2012, 11:51 AM   #1
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Find the limit as x goes to negative infinity for...

x+sqrt(x^2 +2x)

I am pretty sure you multiply by the conjugate x-sqrt(x^2 +2x) / x-sqrt(x^2 +2x) then you get -2x / (x-sqrt(x^2 +2x))... then I think you divide through the numerator and denominator by (1/x) but my answer ends up being (-2 / 0) instead of the correct answer -1 (graphed it in my calculator).

What am I doing wrong?
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May 30th, 2012, 12:28 PM   #2
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Re: Find the limit as x goes to negative infinity for...

Quote:
Originally Posted by soulrain
x+sqrt(x^2 +2x)

I am pretty sure you multiply by the conjugate x-sqrt(x^2 +2x) / x-sqrt(x^2 +2x) then you get -2x / (x-sqrt(x^2 +2x))... then I think you divide through the numerator and denominator by (1/x) but my answer ends up being (-2 / 0) instead of the correct answer -1 (graphed it in my calculator).

What am I doing wrong?
Are you neglecting the fact that x goes to negative infinity? If you divide inside the square root by a negative number, x, it becomes inside the square root. Dividing both numerator and denominator by x gives
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May 30th, 2012, 06:08 PM   #3
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Re: Find the limit as x goes to negative infinity for...

Thanks for your help! I get that x becomes x^2 under the radical but how does going to negative infinity change the denominator from 1 minus sqrt{1+ 2/|x|}) to
1 plus \sqrt{1+ 2/|x|})?

Is it like sqrt( (-1)^2) and you are pulling out the (-1) thus changing the sign?
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May 31st, 2012, 01:14 AM   #4
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Re: Find the limit as x goes to negative infinity for...

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June 1st, 2012, 10:42 AM   #5
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Re: Find the limit as x goes to negative infinity for...

Hi greg thanks for the further help!

I now know what I am totally not getting: the limit of x to - infinity for ( sqrt (x^2 + 2x))/X being = -1

Could you explain how you know that is = -1 when x goes to negative infinity please... I know it does and you are right bu mentally I missing something.
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June 1st, 2012, 11:10 AM   #6
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Re: Find the limit as x goes to negative infinity for...

, not x. is approximately equal to |x| for x a very large negative number so is close to which goes to -1.
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June 1st, 2012, 12:34 PM   #7
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Re: Find the limit as x goes to negative infinity for...

Got it ! Thnks guys so very much!
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January 23rd, 2017, 10:46 AM   #8
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Ive read this thread repeatedly.
Still i dont understand how the sign on the denominator changes to negative: thus resulting in the solution and not in the -2/0 you would instinctualy get.

Please can anyone help.
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January 23rd, 2017, 11:42 AM   #9
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Try writing $y=-x$ so that $y \to +\infty$ as $x \to -\infty$ and thus
$$\lim_{x \to -\infty} x + \sqrt{x^2 + 2x} = \lim_{y \to +\infty} (-y) + \sqrt{(-y)^2 + 2(-y)} = \lim_{y \to +\infty} \sqrt{y^2 - 2y} - y$$
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January 23rd, 2017, 12:47 PM   #10
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Quote:
Originally Posted by miliman13 View Post
Ive read this thread repeatedly.
Still i dont understand how the sign on the denominator changes to negative: thus resulting in the solution and not in the -2/0 you would instinctualy get.

Please can anyone help.
The sign in the denominator does NOT "change" to negative. Since x is going to negative infinity, the denominator, x, is negative.
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