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I'm having troubles showing that $$ \lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = -1. $$ In particular, why is the following derivation wrong? $$ \lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{x\sqrt{1+2/x}}{x} = \lim_{x\rightarrow -\infty} \sqrt{1+2/x} = \sqrt{1+\lim_{x\rightarrow -\infty}(2/x)} = 1.$$

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$\sqrt{x^2}=|x|=-x$ if $x<0$. –  Hans Lundmark Oct 30 '12 at 16:34
    
Thanks! I totally missed that. :) –  somebody Oct 30 '12 at 16:35
    
What you could have done to discover your mistake was to plug in some number like $x=-100000$. –  Hans Lundmark Oct 30 '12 at 16:37
2  
I have trouble with negative numbers. Maybe not for this problem, but for more complicated ones, I would probably let $u=-x$, find the limit of $\dfrac{\sqrt{u^2-2u}}{-u}$ as $u\to\infty$. –  André Nicolas Oct 30 '12 at 16:39
    
@AndréNicolas: I like that approach. –  somebody Oct 30 '12 at 16:42

1 Answer 1

up vote 0 down vote accepted

Your mistake is here $$ \lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{x\sqrt{1+2/x}}{x} = \ldots = 1. $$

The correct is $$ \lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{-x\sqrt{1+2/x}}{x} = \ldots = -1 $$ since $x<0$ ($\sqrt{x^2}=|x|$, for example $\sqrt{(-1254)^2}=1254$).

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