Show that $\lim\limits_{x\longrightarrow0}\frac{1}{x^2}$ does not exist on $\mathbb{R}$

Question

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 :)

Answer 1

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 $\epsilon = 1$.

Answer 2

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 $\delta = \frac{1}{\sqrt M}$.

Answer 3

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 :)