TITLE: Prove that the limit of $\frac{1}{x^2}$ as $x \to 0$ does not exist
QUESTION [0 upvotes]: I am trying to prove this using two different cases. One case shows the limit "$l$" equals some number leads to contradiction and the other one shows that the limit "$l$" is not equal to that same number also leads to contradiction.
I am able to do the first case.
Case 1: let $l=1$. Let $\epsilon=0$. Let $\delta$>0.
Let m be a natural number so that $m > \frac{1}{\delta}$. Then $0<\frac{1}{m}<\delta$.
Let $x=\frac{1}{m}$. This means $0 < x < \delta$.
Also, $f(x)=m^2$ and since $m$ is a natural number, $m^2$ is greater than or equal to $1$.
So $0<|x-0|< \delta$ and yet $|\frac{1}{x^2}-1| = |f(x) - l| \geq 0$. So if $l=1$, this means $1$ is not the limit as $x$ goes to zero of $f(x)$.
For case $2$ where $l\neq 0$, I have a possible solution, but I think I may have let certain numbers depend on other numbers in a way that does not work.
Case 2: $l$ is not $1$. Let $\epsilon=-l$. Let $\delta>0$.
Let $m$ be a natural number so that $\frac{1}{\delta} < m < \sqrt{2l}$.
Then $1 < \delta m < \delta\sqrt{2l}$. So $0 < \frac{1}{m} < \delta < \delta\sqrt{2l}$. (I know that $1/m$ is greater than zero since m is a natural number).
Let $x = 1/m$. Now we have $0 < x < \delta$.
$f(x) = m^2$. $f(x) < 2l$ since $m < \sqrt{2l}$.
Notice $2l = l - \epsilon$ and also that $0 = l + \epsilon$ So, since $m>0$, we have $0 < f(x) < 2l$ which gives us $l + \epsilon < f(x) < l - \epsilon$. Rearranging this inequality I can get $|f(x) - l|> \epsilon$
So if $l$ is not equal to zero, $l$ is not the limit $f(x)$ as $x$ goes to zero

REPLY [1 votes]: It is easy to work in the following way: 
Suppose $1/x^2\to L$ as $x\to 0$. By the limit laws, $x^2\cdot 1/x^2\to 0\cdot L=0$, as $x\to 0$, but  $(x^2\cdot 1/x^2)=1\to 1$ and by uniqueness of the limit $1=0$, contradiction.
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