Is it valid to write $\displaystyle \lim_{x \to 0} \frac{1}{x^2} = \infty$? AFAIK the limit of a term does not exist if that term does not converge, but I haven't found a suiting question here yet.  This probably is a double of a similar question.
 A: Assigning a value to divergent limits is not entirely standardized among mathematicians.
One reasonable standard is that $\infty$ is not a number, while we want all limits to be numbers, so $\lim_{x\to 0}\frac{1}{x^2}$ simply doesn't exist.
Another standard, probably more common, tries to distinguish functions that grow arbitrarily large, from limits that behave erratically and don't converge.  Then $\lim_{x\to 0}\frac{1}{x^2}=+\infty$.  The $+$ is necessary to distinguish this case from its negative.
A: Yes, $\lim_{x\rightarrow 0 } f(x) = \infty$ means that $f(x)$ diverges to positive infinity as $x$ tends to $0$.
A: It is valid to write $\lim_{x \to 0} \frac{1}{x^2} = \infty$. However, this merely notation for the following: For all $M \in \mathbf{R}$ there is $\delta \in \mathbf{R}$ such that if $0<|x|<\delta$ then $\frac{1}{x^2} > M$, essentially the function $f(x)=\frac{1}{x^2}$ grows without bound as $x$ approaches $0$.
A: $\infty$ is a special case, but the notation is standard. The statement  
$$\lim_{x\to a} f(x) = \infty$$
means that the closer $x$ approaches $a$, the larger $f(x)$ gets. In symbols:
For all $M > 0$ there exists $\delta > 0$ such that $0 < |x-a| < \delta \implies f(x)>M$.
