Is my proof of the continuity of $\frac{1}{x}$ correct? I need to prove that $\frac{1}{x}$ is continuous, but my proof, as far as I did now, only applies to the interval $x\ge1$ wich is a progress. So, there's how I did:
$$\frac{|x-a|}{|x|}<|x-a|<\delta \tag{for $x>1$}$$
Then, by choosing $\delta=\epsilon|a|$ we have:
$$\frac{|x-a|}{|x|}<\epsilon|a|\implies\frac{|x-a|}{|x||a|}<\epsilon\implies\frac{|x-a|}{|xa|}<\epsilon \implies \frac{|a-x|}{|xa|}<\epsilon \implies \left|\frac{1}{x} - \frac{1}{a}\right|<\epsilon$$
This, implies that $\lim \limits_{x\to a}\frac{1}{x} = \frac{1}{a}$ for $x>1$. I did not think so much about the $x\le1$ case, but I would like if this proof is correct. 
 A: There are a few not very big things wrong with the proof, and a major problem: you do not explain the logic of the argument.  
Let $a\ne 0$, and let $f(x)=\frac{1}{x}$. We want to show that $f$  is continuous at $x=a$.  In what follows, we will assume that $a\gt 0$. By putting in the appropriate absolute value signs, one can transform the proof into one that works for all $a\ne 0$.
We want to show that for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $|x-a|\lt \delta$, then $|f(x)-f(a)\lt \epsilon$.  
If following the argument poses difficulty, let $a=0.4$.  Now let $\epsilon \gt 0$ be given.
It is easy to see that for any $x\ne 0$,  we have $|f(x)-f(a)|=\frac{|x-a|}{a|x|}$.
We have control over $|x-a|$, since we have great freedom in selecting $\delta$. The number $a$ is fixed. We need to control the size of $x$, since if $x$ is close to $0$, the $x$ in the denominator may make $|f(x)-f(a)|$ big.
Now comes the first key step. We will make sure ultimately that $\delta\le \frac{a}{2}$. Why? Well, then if $|x-a|\le \delta$, then we will have $\frac{a}{2}\lt x\lt \frac{3a}{2}$. The important part is that we will have $x\gt \frac{a}{2}$, so $x$ will stay nicely away from  $0$.
Thus if $\delta\le \frac{a}{2}$, and $|x-a|\lt \delta$, we have
$$|f(x)-f(a)|\lt \frac{2}{a^2}|x-a|.$$
Now to make sure that $|f(x)-f(a)|\lt \epsilon$, it is enough to make sure, in addition to the $\delta \le \frac{a}{2}$, that $\delta\le \frac{a^2\epsilon}{2}$.
Finally, we choose our $\delta$: it is $\min\left(\frac{a}{2},\frac{a^2\epsilon}{2}\right)$. By our arguments above, if $|x-a|\lt \delta$, then $|f(x)-f(a)|\lt \epsilon$.
Remark: The restriction to $x\ge 1$ in the argument was an attempt to avoid small denominators, so you were aware of the small denominators problem. In your proof, we can make sure that $x\ge 1$ if $a\gt 1$ by picking $\delta$ small enough, for example $\delta\le \frac{a-1}{2}$. But you did not do that.
There is a problem with the assertion that if $\delta=\epsilon |a|$ then everything is OK. Imagine, to be  silly, that $a=2$ and $\epsilon=47$. (This is the silly part.) Then $x$ being within $\epsilon |a|$ of $a$ guarantees nothing,  $x$ could even be equal to $0$. 
You may think the argument of the answer tedious, and overkill. True, and true. But I wanted to describe the issues in detail, since the basic pattern is fairly standard. 
A: You are correct.And for cases where x is less then or equal to 1 function is not continuous,since it is not continuous at 0.
We can see that left-sided and right sided limit at 0 are not equal,thus it does not exist,thus function is not continuous for $x \le 1$
