Show that $\lim\limits_{x\longrightarrow0}\frac{1}{x^2}$ does not exist on $\mathbb{R}$ Show that $\lim\limits_{x\longrightarrow0}\frac{1}{x^2}$ does not exist on $\mathbb{R}$
My attempt:
In order for the limit to exist, Cauchys convergence criterium has to hold:
$\forall \epsilon > 0\,\,\, \exists \delta > 0:\forall x,y \in \dot{\mathcal{U}}_\delta(0)\cap \mathbb{R} \Longrightarrow |\frac{1}{x^2}-\frac{1}{y^2}|<\epsilon$
$|\frac{1}{x^2}-\frac{1}{y^2}|<\epsilon \Longleftrightarrow |\frac{y^2-x^2}{(xy)^2}|<\epsilon \Longleftrightarrow |y-x||y+x|<\epsilon(xy)^2$
Now since $x,y \in \dot{\mathcal{U}}_\delta(0)\cap \mathbb{R}$ we could choose $d(x,y)=|y-x|=\frac{\delta}{2}$
This would mean:
$\frac{\delta}{2}|y+x|<\epsilon(xy)^2$
Now let $x \longrightarrow 0$
To keep the distance $d(x,y)=\frac{\delta}{2}$ we now set $y:=\frac{\delta}{2}$
$\frac{\delta}{2}|\frac{\delta}{2}+0|<\epsilon(0*\frac{\delta}{2})^2$
This would require $\delta<0$ in order for our inequation to hold for any $\epsilon > 0$
Which is a contradiction.
It would be very helpful if someone could give me some feedback :)
 A: Cauchy Criterium for function $f$ in point $a$:
$$\forall \epsilon >0, \exists \delta >0, \forall x \in \left| x- a \right|< \delta, \space  \forall y \in \left| y- a \right|< \delta, \space \left|f(x) - f(y) \right|< \epsilon$$
negation:
$$\exists \epsilon >0, \forall \delta >0, \exists x \in \left| x- a \right|< \delta, \space  \exists y \in \left| y- a \right|< \delta, \space \left|f(x) - f(y) \right|\geqslant \epsilon$$
So, really, $x$ and $y$ are functions of $\delta$. It is enough to find sequences $\left\lbrace x_n\right\rbrace \rightarrow a$ and $ \left\lbrace y_n\right\rbrace \rightarrow a$, for which held $\left|f(x_n) - f(y_n) \right|\geqslant \epsilon$.
Now let's take $x_n = \frac{1}{\sqrt n}$ and $y_n = \frac{1}{\sqrt {n+1}}$. Both $\rightarrow 0$. But then we have
$$\left|f(x_n) - f(y_n) \right| = \left|  n- (n+1) \right| = 1$$
So we find sequences for $\epsilon = 1$.
A: We can just proof that 
$$
\lim_{x \to 0} \frac{1}{x^2}=+\infty
$$
By definition this is equivalent to show that
$$
\forall M > 0 \  \exists \ \delta > 0 : \forall x \in (-\delta,\delta)\quad  \text{ we have} \quad \frac{1}{x^2} > M
$$
Let fix $M >0$, we are looking for the $\delta$, hence
$$
\frac{1}{x^2} > M \iff x^2 < \frac{1}{M} \iff \frac{1}{\sqrt M} < x < \frac{1}{\sqrt M}
$$
So we can just choose $\delta = \frac{1}{\sqrt M}$.
A: Okey so if I want to show the same for $f(x)=\ln(x)$
I choose $(x_n) $ with $x_n=\frac{1}{n}$
and $(y_n) $ with $y_n=\frac{1}{2n}$, 
$y_n\longrightarrow0$ and $x_n\longrightarrow0$
Now $|f(x_n)-f(y_n)|=|\ln(\frac{1}{n})-\ln(\frac{1}{2n})|=|\ln(\frac{\frac{1}{n}}{\frac{1}{2n}})|=|\ln(\frac{2n}{n})|=|\ln(2)|$
so we find two sequences we can use to show the negation of the cauchy criterium holds for $\epsilon=\ln(2)$
If this is correct, you did me an amazing favour because you also opend some eyes in terms of analysis in general :)
