Hello I was hoping that someone might be able to verify that the following proof that $\lim_{x\to 0} {1\over x}$ does not exist is correct.
First assume that $\lim_{x\to 0} {1\over x} = L$. This means that for every $\epsilon > 0$ there is a $\delta > 0$ such that for all $x$ if $0 < \lvert x\rvert < \delta$ then $\lvert{1\over x} – L \rvert <\epsilon.$ Now let $\epsilon=1$ and choose $x < \min(1,\delta)$. Therefore there is a $\delta>0$ such that if $0 < \lvert x\rvert < \delta$ then $\lvert{1\over x} – L \rvert <\epsilon$. It follows that $\lvert{1\over x}\rvert < 1+ \lvert L\rvert$. However clearly $\lvert{1\over x}\rvert>1$ and therefore a contradiction has been reached and there is no number $L$ such that $\lim_{x\to 0} {1\over x} = L$.
Best Answer
You could argue with the equivalent definition of sequences:
Assume it exists a $L \in \mathbb{R}$ with $\lim_{x \to 0} \frac{1}{x}= L$ then for any sequence $a_n \to 0 $ with $a_n \neq 0$ it follows $\frac{1}{a_n} \to L$.
Now take $a_n = \frac{1}{n}$, then $\frac{1}{a_n} = n \not \to L$ for any $L\in \mathbb{R}$, since the sequence is unbounded (any convergent sequence is bounded).