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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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 $\sqrt{(-1254)^2}=1254$).