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Working through some limit exercises. The answer sheet says the limit below does not exist. Is this correct. Shouldn't it be $-\infty$? $$\lim_{x \to 0^+} \left( \frac{1}{\sqrt{x^2+1}} - \frac{1}{x} \right)\ \ \ $$

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Are you sure the question wasn't about just $\lim_{x\to 0}$? As it is, you're right. –  tomasz Jun 15 '13 at 0:17
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Perhaps the book means it doesn't exist finitely... –  DonAntonio Jun 15 '13 at 0:17
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4 Answers

up vote 7 down vote accepted

It is true that as $x$ approaches through positive values, $\frac{1}{\sqrt{1+x^2}}-\frac{1}{x}$ becomes large negative. However, some people do not allow $\infty$ or $-\infty$ as answers to a limit problem. As a simpler example, some would say that $\lim_{x\to\infty}x^2$ does not exist, and some would say $\lim_{x\to\infty} x^2=\infty$.

Remark: Mathematical English has dialects. When you answer a question, it may be necessary to conform to the local dialect. (On a test, I would accept either answer if proper justification were given, but cannot guarantee that someone else would.)

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+1 Love the remark about dialects. It is a pain to deal with it ... –  Calvin Lin Jun 15 '13 at 0:50
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The same book states that $lim_{x \to \left(\frac{\pi^-}{2}\right)} \left(\tan x\right)^x = \infty$ –  hondaman Jun 15 '13 at 1:16
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$hondaman: Then they are being very inconsistent, and one one of the answers is incorrect. –  André Nicolas Jun 15 '13 at 1:21
    
@hondaman: See the reply to your comment (if you haven't already). –  Cameron Buie Jun 15 '13 at 3:50
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$$\lim_{x \to 0^+} \left( \frac{1}{\sqrt{x^2+1}} - \frac{1}{x} \right)=\lim_{x\to 0^+}\frac{x-\sqrt{x^2+1}}{x\sqrt{x^2+1}}=$$

$$=-\lim_{x\to 0^+}\frac1{x\sqrt{x^2+1}(x+\sqrt{x^2+1})}=\lim_{x\to 0^+}-\frac1{x^2\sqrt{x^2+1}+x^3+x}=-\infty$$

so you're right...unless the book meant "doesn't exists finitely " , say.

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That is correct. Since $-\infty$ is not a real number, the limit does not exist.

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since $$\lim_{x \to 0^+} \frac{1}{\sqrt{x^2+1}} =1$$ and $$\lim_{x \to 0^+} = - \frac{1}{x} =-\infty$$ hence you can say the given limit is $-\infty$ or the limit doesn't exist since it's not finite.

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