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 xsqrt(x^2 +2x) / xsqrt(x^2 +2x) then you get 2x / (xsqrt(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  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Find the limit as x goes to negative infinity for... Quote:
 
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... 
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... , not x. is approximately equal to x for x a very large negative number so is close to 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 $$\lim_{x \to \infty} x + \sqrt{x^2 + 2x} = \lim_{y \to +\infty} (y) + \sqrt{(y)^2 + 2(y)} = \lim_{y \to +\infty} \sqrt{y^2  2y}  y$$ 
January 23rd, 2017, 12:47 PM  #10 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  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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