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 $-\infty$?
$$\lim_{x \to 0^+} \left( \frac{1}{\sqrt{x^2+1}} - \frac{1}{x} \right)\ \ \ $$
 A: 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.) 
A: $$\lim_{x \to 0^+} \left( \frac{1}{\sqrt{x^2+1}} - \frac{1}{x} \right)=\lim_{x\to 0^+}\frac{x-\sqrt{x^2+1}}{x\sqrt{x^2+1}}=$$
$$=-\lim_{x\to 0^+}\frac1{x\sqrt{x^2+1}(x+\sqrt{x^2+1})}=\lim_{x\to 0^+}-\frac1{x^2\sqrt{x^2+1}+x^3+x}=-\infty$$
so you're right...unless the book meant "doesn't exists finitely "
, say.
A: 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.
A: That is correct. Since $-\infty$ is not a real number, the limit does not exist.
