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 May 30th, 2012, 11:51 AM #1 Senior Member   Joined: Jan 2012 Posts: 159 Thanks: 0 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?
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 $x^2$ inside the square root. Dividing both numerator and denominator by x gives $-\frac{2}{1+ \sqrt{1+ \frac{2}{|x|}}}$

 May 30th, 2012, 06:08 PM #3 Senior Member   Joined: Jan 2012 Posts: 159 Thanks: 0 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?
 May 31st, 2012, 01:14 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Find the limit as x goes to negative infinity for... \begin{align*}\lim_{x\to-\infty}\,x\,+\,\sqrt{x^2\,+\,2x}\,&=\,\lim_{x\to-\infty}\,$$\(x\,+\,\sqrt{x^2\,+\,2x}$$\,\cdot\,\fr ac{x\,-\,\sqrt{x^2\,+\,2x}}{x\,-\,\sqrt{x^2\,+\,2x}}\) \\ &=\,\lim_{x\to-\infty}\,\frac{-2x}{x\,-\,\sqrt{x^2\,+\,2x}} \\ &=\,\lim_{x\to-\infty}\frac{2x}{\sqrt{x^2\,+\,2x}\,-\,x} \\ &=\,\lim_{x\to-\infty}\,\frac{2}{\frac{\sqrt{x^2\,+\,2x}}{x}\,-\,1} \\ &=\,\frac{2}{-1\,-\,1}\,=\,-1\end{align*}
 June 1st, 2012, 10:42 AM #5 Senior Member   Joined: Jan 2012 Posts: 159 Thanks: 0 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.
 June 1st, 2012, 11:10 AM #6 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Find the limit as x goes to negative infinity for... $\sqrt{x^2}= |x|$, not x. $\sqrt{x^2+ x}$ is approximately equal to |x| for x a very large negative number so $\frac{\sqrt{x^2+ x}}{x}$ is close to $\frac{|x|}{x}$ which goes to -1.
 June 1st, 2012, 12:34 PM #7 Senior Member   Joined: Jan 2012 Posts: 159 Thanks: 0 Re: Find the limit as x goes to negative infinity for... Got it ! Thnks guys so very much!
 January 23rd, 2017, 10:46 AM #8 Newbie   Joined: Jan 2017 From: FL Posts: 12 Thanks: 0 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.
 January 23rd, 2017, 11:42 AM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Try writing $y=-x$ so that $y \to +\infty$ as $x \to -\infty$ and thus $$\lim_{x \to -\infty} x + \sqrt{x^2 + 2x} = \lim_{y \to +\infty} (-y) + \sqrt{(-y)^2 + 2(-y)} = \lim_{y \to +\infty} \sqrt{y^2 - 2y} - y$$ Thanks from miliman13
January 23rd, 2017, 12:47 PM   #10
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Quote:
 Originally Posted by miliman13 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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