# Why does this limit not exist?

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)\ \ \$$

• 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
• Perhaps the book means it doesn't exist finitely... – DonAntonio Jun 15 '13 at 0:17

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 $\lim_{x\to\infty} x^2=\infty$.
• 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
• $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 $$\lim_{x \to 0^+} \left( \frac{1}{\sqrt{x^2+1}} - \frac{1}{x} \right)=\lim_{x\to 0^+}\frac{x-\sqrt{x^2+1}}{x\sqrt{x^2+1}}=$$ $$=-\lim_{x\to 0^+}\frac1{x\sqrt{x^2+1}(x+\sqrt{x^2+1})}=\lim_{x\to 0^+}-\frac1{x^2\sqrt{x^2+1}+x^3+x}=-\infty$$ so you're right...unless the book meant "doesn't exists finitely " , say. 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. • Needs an upvote! – Namaste May 30 '14 at 16:52 That is correct. Since$-\infty\$ is not a real number, the limit does not exist.