Does $\lim_{x \to 0} \frac{1}{x^2} = \infty ?$ I was curious as to whether  $$\lim_{x \to 0}\frac{1}{x^2}=\infty $$
Or the limit does not exist? Because doesn't a limit exist if and only if the limit tends to a finite number?  
 A: Note that for any $B>0$, $\frac{1}{x^2}>B$ whenever $0<|x|<\frac{1}{\sqrt{B}}$. By definition the limit is $\infty$.
A: Recall that a limit exists if the left and right limits exist and agree. Just to clarify, yes, $\infty$ or $-\infty$ are suitable limits.
Since $(-x)^2=x^2,$ then 
$$\displaystyle\lim_{x\to0^-}\dfrac{1}{x^2}=\displaystyle\lim_{x\to0^+}\dfrac{1}{x^2}=\infty,$$
and so you are correct, 
$$\displaystyle\lim_{x\to0}\dfrac{1}{x^2}=\infty.$$
A: Infinity is a special case. We write $$\lim_{x\to a}f(x)=\infty$$ to mean:
For all $M\in\mathbb R$ there exists $\delta > 0$ such that $0<|x-a|<\delta\implies f(x)>M$.
As you noticed, this is different from the definition used for a finite limit.
A: "Because doesn't a limit exist if and only if the limit tends to a finite number? "
It's a matter of notation and/or definition.  
I'd argue "The limit is infity" and "the limit does not exist" are NOT contradictory statements:
.....
The symbols $\lim\limits_{x \rightarrow a} f(x) = \infty$ means, by definition, for every $\epsilon > 0$ we can find an $M$ so that whenever $|x-a| < \epsilon$, it follows $f(x) > M$. (Colloquially, "as $x$ gets close to $a$, $f(x)$ gets arbitrarily large".) 
[Equivalently, for any $M$ no matter how large we can find an $\epsilon > 0$ so that for all $|x - a| < \epsilon$ we will have $f(x) > M$.  That might be a more direct interpretation of our intuitive understanding.]
And that is true for the symbols you wrote. 
When we write $\lim\limits_{x \rightarrow a} f(x) = L$ and $L$ is a finite value, we mean, for every $\epsilon > 0$ we can find an $\delta$ so that whenever $|x-a| < \epsilon$ $|f(x) -L| < \delta$. (Colloquially, "as $x$ gets close to $a$, $f(x)$ gets close to $L$").
Notice neither of these have to be true.  $\lim\limits_{x\rightarrow 0} \sin \frac 1x \ne \pm \infty; \lim\limits_{x\rightarrow 0} \sin \frac 1x \ne L$ for any finite $L$.
If there exist such a number we call such a number "the limit as $x\rightarrow a$".  So "the limit is $L$" is meaningful and well defined.
If $\lim\limits_{x\rightarrow a} f(x) = \infty$ it would be true to say "No finite limit exists". It'd even be true to say "No limit exists" as it is understood that a limit is a finite number.
However the sounds "The limit is infinity" are sounds we can produce by exhaling air through our vocal cords. And we can declare:  "That means $\lim\limits_{x\rightarrow a} f(x) = \infty$".
And that would be just fine.
So the two statements "No limit exists" and "The limit is infinity" are not necessarily inconsistent. Although for personal style you'll probably want to pick one and stick to it.  It should be very clear in context what you mean.  I, personally would use "no limit exists" to mean that neither $\lim f(x) = \pm \infty$ nor $\lim f(x) = L$ I'd say "The function diverges to positive infinity" for $\lim f(x) = \infty$.
.....
At any event.... for any $\epsilon > 0$ there will be an $M$ so that $|x| < \epsilon \implies \frac 1{x^2} > M$. (or for any $M$ we can find an $\epsilon > 0$).
That is true and I believe you understand and thoroughly accept that.
A: good question. if you have $$\lim_{x \to 0} \frac{1}{x^2}$$
as $x \to 0$, the denominator gets closer to 0, and the number gets bigger. example:
$$\frac {1}{0.01^2}=10000$$
$$\frac {1}{0.001^2}=1000000$$
$$\frac {1}{0.00001^2}=100000000$$
BUT(and this is a big but)  you cannot have division by $0$, but it can be something very very close to it.
you can also have any real number! any number getting to 0 will register as $+\infty$
BUT(yes, another but!) if you use a POSITIVE imaginary number(i.e 4+5i), you'll get to $-\infty$  according to my problem on negative imaginary numbers, $-i=i^3$ thus $$(-i^2)=(i^3)^2=i^6=i^2=\sqrt {-1}$$  so a negative imaginary number (i.e. 9-4i)is also tends to $-\infty$
(yet another!)but, i did some more trial and error and found out something else important: in $a+bi$, if $a>b>0$, then the answer will be positive. case in point:
$$[6+4i]^2=[36-16]=20$$
and if you have $a-bi$ and $a<0>b$, then the answer is positive. example:
$$[4-6i]^2=[16-[-36]]=[16+36]=52$$
or
$$[16-3i]^2=[256-[-9]]=[256+9]=265$$
so you could also say that $$\left(\lim_{a \to 0} > \lim_{b \to 0}\right) \frac {1}{a+bi}\to \infty$$
and either $a$ or $b$ can be 0, but NOT both.
so you can rewrite the equation for this 
