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? $$ \lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{x\sqrt{1+2/x}}{x} = \lim_{x\rightarrow -\infty} \sqrt{1+2/x} = \sqrt{1+\lim_{x\rightarrow -\infty}(2/x)} = 1.$$
Determining a limit
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calculus
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4$\sqrt{x^2}=|x|=-x$ if $x<0$. – 2012-10-30
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0Thanks! I totally missed that. :) – 2012-10-30
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0What you could have done to discover your mistake was to plug in some number like $x=-100000$. – 2012-10-30
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2I 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 $u\to\infty$. – 2012-10-30
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0@AndréNicolas: I like that approach. – 2012-10-30
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0@somebody : +1 for showing your work! – 2012-10-30
1 Answers
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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 $\sqrt{(-1254)^2}=1254$).