Prove that the limit $\lim_{n\to\infty} \frac{\alpha + n}{\beta + n} $ equals $1$ I'd like to prove that
$$\lim_{n\to\infty} \frac{\alpha + n}{\beta + n} = 1$$
I've worked out that $$\frac{|\alpha - \beta|}{|\beta + n|} < \epsilon$$ if $n < N$ for integers $n$ and $N$, but I'm not sure where to go from there.
 A: write $$\frac{\frac{\alpha}{n}+1}{\frac{\beta}{n}+1}$$ and this tends to $1$ for $n$ tends to infinity
A: If $\alpha = \beta$ it is obvious. Otherwise, fix $\varepsilon > 0$ and look for $N \in \mathbb{N}$ such that 
$$\Big|\frac{\alpha+n}{\beta+n}-1\Big| < \varepsilon \text{ when } n > N. $$
You find immediately
$$\Big|\frac{\alpha-\beta}{\beta+n} \Big| < \varepsilon \rightarrow  \frac{|\alpha-\beta|}{\varepsilon} < |\beta+n| \leq |\beta|+n$$
So if you set $N > \frac{|\alpha-\beta|}{\varepsilon}-|\beta|$ you get the result.
A: Note that
$$
\frac{\alpha + n}{\beta + n} = \frac{\alpha - \beta + \beta + n}{\beta + n}\\
= \frac{\alpha - \beta}{\beta + n} + \frac{\beta + n}{\beta + n}\\
= \frac{\alpha - \beta}{\beta + n} + 1
$$
so by definition $\lim_{n\to \infty}\frac{\alpha + n}{\beta + n} = 1$ means that for any $\epsilon>0$, there is an $N\in \Bbb N$ such that as long as $n>N$, we have
$$
\left|\frac{\alpha + n}{\beta + n} - 1\right|<\epsilon
$$
However, by the above argument, this is inequality is equivalent to
$$
\left|\frac{\alpha - \beta}{\beta + n}\right|<\epsilon
$$
and you already know that there is an $N$ that works in this case. So we're done.
A: You need to establish
$$|\alpha-\beta|<|\beta+n|\epsilon.$$
This is obviously obtained with
$$n>\frac{|\alpha-\beta|}\epsilon-\beta,$$ where the RHS is a well-defined finite number.
A: Note that
$$
\left|\frac{\alpha+n}{\beta+n}-1\right|
= \left|\frac{\alpha+n-\beta-n}{\beta+n}\right|
= \left|\frac{\alpha-\beta}{\beta+n}\right|
= \frac{|\alpha-\beta|}{|\beta+n|}
$$
You showed that the right side is finally $<\epsilon$ for large enough $n$. This means that the left side will too. And this is the definition of
$$\lim_{n\to\infty} \frac{\alpha+n}{\beta+n}=1.$$
A: If $\alpha=\beta$ it's trivial if not you can divide by $|\alpha-\beta|$ to get
$$\frac1{|\beta+n|} <\frac\epsilon{|\alpha-\beta|} $$
Now take the greatest integer smaller or equal to $\beta+n$ denote it by $[\beta+n] $ then $\frac1{|\beta+n|} <\frac1{[|\beta+n|]} <\frac\epsilon{|\alpha-\beta|} $ the last inequality is due the Archimedian property. 
