1. ## Question on notation

Prove $\displaystyle \lim_{x \to 0}\frac{1}{x}$ does not exist.

Proof:
Asssume, to the contrary, that $\displaystyle \lim_{x \to 0}\frac{1}{x}$ exists. Then there exists a real number L such that $\displaystyle \lim_{x \to 0}\frac{1}{x}=L$. Let $\displaystyle \epsilon = 1$. Choose an integer $\displaystyle n$ such that $\displaystyle n>\lceil 1/\delta \rceil \geq 1$. Since $\displaystyle n>1/\epsilon$, it follows that $\displaystyle 0<1/n<\delta$. ...........

Question: In sequence $\displaystyle n$ denotes the term of the sequence. What does $\displaystyle n$ denote in this context?

2. I don't think your proof is quite what's necessary. I think you should show that

$\displaystyle \displaystyle{\lim_{x\to 0^{+}}\frac{1}{x}=\infty.$

Different books sometimes define things differently, but if you can prove that one of the one-sided limits is infinite, then the limit does not exist. So how would you go about proving this? The definition of this limit is as follows:

$\displaystyle \displaystyle{\lim_{x\to a^{+}}f(x)=\infty$ if and only if

$\displaystyle \displaystyle{(\forall N>0)(\exists\delta>0)(0<x-a<\delta\Rightarrow f(x)>N)}.$

So, you need to prove that

$\displaystyle \displaystyle{{(\forall N>0)(\exists\delta>0)\left(0<x<\delta\Rightarrow \frac{1}{x}>N\right)}.$

In answer to your sequence question, in this context, n is just a number you've defined in your proof.