Showing that the $\lim_{x\to 0}\frac{1}{x^2}$ does not exist I want to show that the $\lim_{x\to 0}\frac{1}{x^2}$ does not exist.
I start with $|x-0|<\delta$ and $|\frac{1}{x^2}|<\epsilon$
I let $\epsilon = 1$ so $|\frac{1}{x^2}|<1$
Then $|1|<x^2$ which means $|1|<x$
If I make my $\delta= 1$
$|x-0|<1$ 
and $|x|<1$
I can't have $|x|<1$ and $|1|<x$ so I can't have a limit here. Is that right? 
 A: Suppose that the limit exists and equals $c\in\mathbb{R}$. 
Then for e.g.
$\epsilon>1$ some $\delta>0$ must exist with $\left|x\right|<\delta\implies\left|\frac{1}{x^{2}}-c\right|<1$.
However, if we take $\left|x\right|$ small enough then $\left|\frac{1}{x^{2}}-c\right|$
will definitely exceed $1$ (do you see why?). 
We conclude that the limit does not exist.
A: Actually the limit exists and is equal to $\infty$. No matter from which direction you approach $0$ (from negative values or from positive values) the term $x^2$ is positive. Moreover as $x \to 0$, the term $x^2$ becomes very small, so that $\frac1{x^2}$ becomes very big (in other words, it grows steadily to infinity). Bringing all these together
$$\lim_{x\to 0-}\frac1{x^2}=\lim_{x\to 0+}\frac{1}{x^2}=+\infty$$
A: By definition, the limit of a function at a point is defined as
$$\exists L:\left( {\forall \varepsilon  > 0,\exists \delta  > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta  \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$
and its negation will be (see this post)
$$\forall L,\exists \varepsilon  > 0:\left( \forall \delta  > 0,\exists x:(0 < \left| x - a \right| < \delta  \wedge |f(x) - L| \ge \varepsilon \right))$$
So for your special case we shall prove that
$$\forall L,\exists \varepsilon  > 0: ( \forall \delta  > 0,\exists x:(0 < \left| x  \right| < \delta  \wedge |\frac{1}{x^2} - L| \ge \varepsilon )$$
You can see this as a game! Your opponent chooses $L$ then you select an $\epsilon$. Next, the opponent chooses $\delta$ and then you choose the $x$. You should be wise enough to make proper choices so that you will win the game no matter what your opponent chooses. So your choices must satisfy the conditions $\left| x  \right| < \delta  \wedge |\frac{1}{x^2} - L| \ge \varepsilon$ regardless of opponent's choices. So, let us play!
The opponent makes a choice for $L$. we choose $\varepsilon$ to be $1$. Then the opponent selects a $\delta$ and we should make our final choice for $x$ such that
$$\left| x  \right| < \delta  \wedge |\frac{1}{x^2} - L| \ge 1$$
we can choose $x_n=\frac{1}{\sqrt{n+L}}$ and our conditions turns to be
$$\left| \frac{1}{\sqrt{n+L}}  \right| < \delta  \wedge |n| \ge 1$$
So, no matter what the opponent chooses, we will select large enough values for $n$ such that the above conditions are satisfied and we win the game! Hence, the limit does not exist!
A: Find the following limit:
$$\lim _{x \rightarrow 0} \frac{1}{x^{2}}$$
$$\lim _{x \rightarrow 0} \frac{1}{x^{2}}=\lim _{x \rightarrow 0} \exp \left(\log \left(\frac{1}{x^{2}}\right)\right):$$
$$\lim _{x \rightarrow 0} \exp \left(\log \left(\frac{1}{x^{2}}\right)\right)$$
$$\exp \left(\log \left(\frac{1}{x^{2}}\right)\right)=\exp (-2 \log (x)):$$
$$\lim _{x \rightarrow 0} \exp (-2 \log (x))$$
$$\lim _{x \rightarrow 0} \exp (-2 \log (x))=\exp \left(\lim _{x \rightarrow 0}-2 \log (x)\right):$$
$$\exp \left(\lim _{x \rightarrow 0}-2 \log (x)\right)$$
Applying the product rule, write $\lim _{x \rightarrow 0}-2 \log (x)$ as $-2 \lim _{x \rightarrow 0} \log (x)$ :
$$\exp \left(-2 \lim _{x \rightarrow 0} \log (x)\right)$$
$$\lim _{x \rightarrow 0} \log (x)=-\infty:$$
$$e^{-2[-\infty}$$
$$e^{-2(-\infty)}=\infty:$$
$$\lim_{x\to 0}\frac{1}{x^2}:
\infty$$
