Elementary limit proof I am given
$\lim_{n\to\infty}{x_n}=1$
And have to prove
$\lim_{n\to\infty}{\frac{2+x_n^2}{x_n}}=3$
Which is very obvious, but I have to prove it using only the definition of the limit:
$\lim_{n\to\infty}{x_n}=a$ iff for every $\alpha>0$ there exists an $N$ such that $n>N$ implies $|x_n-a|<\alpha$
Now I started with the fact that we are asked to prove that $|\frac{2+x_n^2}{x_n} - 3|$ can be made arbitrarily small. Rewriting: $|\frac{x_n^2-3x_n+2}{x_n}| = |\frac{(x_n-1)(x_n-2)}{x_n}| = |x_n-1||x_n-2||x_n^{-1}|$
Now we can (I think) say:
Let $N_\epsilon$ be such that $n>N_\epsilon$ implies $|x_n-1|<\epsilon$ for all $\epsilon>0$ 
Than we know that 
$|x_n-1||x_n-2||x_n^{-1}| < \epsilon(\epsilon+1)|x_n^{-1}|$ for all $n>N_\epsilon$
And now I was a bit stuck, and tried this:
We know that there exists an $N_{\frac{1}{2}}$ such that $n>N_\frac{1}{2}$ implies $|x_n-1|<\frac{1}{2}$ 
This means $x_n > \frac{1}{2}$ for all $n>N_\frac{1}{2}$. Which in turn means $0<x_n^{-1}<2$ for all $n>N_\frac{1}{2}$
So now we know(?)
$|x_n-1||x_n-2||x_n^{-1}| < \epsilon(\epsilon+1)2$ for all $n>\max{N_\epsilon,N_{\frac{1}{2}}}$
And because $\epsilon(\epsilon+1)2$ can take on any value $>0$ we have shown that  $\lim_{n\to\infty}{\frac{2+x_n^2}{x_n}}=3$
Now this seems VERY ugly and long and just not the way this is supposed to be done. Can someone please tell me A: if this proof is even correct, and B: a nicer way to prove this (remember it MUST be done using only the above definition of a limit and elementary absolute value stuff)
Thanks
 A: Try rewriting $$\frac {2+x_n^2}{x_n}= \frac {2}{x_n}+ \frac{x_n^2}{x_n} $$
Let $|x_n-1|< \epsilon$. Then $ \frac {2}{1+\epsilon}<\frac {2}{x_n} < \frac {2}{1-\epsilon}$
A: You can use simply arithmetic of limit:
$\lim_{n \to \infty} x_n=1$, so $\lim_{n \to \infty} (x_n)^2=(\lim_{n \to \infty} x_n)^2=1 \times 1$.
Next $\lim_{n \to \infty} x_n^2+2=3$. Finally:
$\lim_{n \to \infty} \frac{x_n^2+2}{x_n}=\frac{lim_{n \to \infty} x_n^2+2}{\lim_{n \to \infty} x_n}=\frac{3}{1}=3$
A: Unfortunately, proofs with this definition can be fairly tedious and long-winded for all but the most simple sequences. The only way I could think of simplifying it is as follows:
As before, we have $$\left|\frac{2+x_n^2}{x_n}-3\right| = |x_n-1|\left|1-\frac{2}{x_n}\right|$$
We can find $N$ such that $|x_n-1| < \frac{\epsilon}{3}$ and $|x_n-1| < \frac{1}{2}$ for $n > N$. From there, we have that $\left|1-\frac{2}{x_n}\right| < 3$ so
$$n > N \implies |x_n-1|\left|1-\frac{2}{x_n}\right| < 3\frac{\epsilon}{3} = \epsilon$$
This isn't a particularly pretty definition for a limit: the topological formulation (while entirely equivalent) is, in my opinion, much nicer.
