Proving the general limit $\lim_{n\to\infty} \frac{an+c}{bn+d} = \frac{a}{b}$ I'm trying to prove the general 

$$ \lim_{n\to\infty} \frac{an+c}{bn+d} = \frac{a}{b} $$

using the definition of the limit of a sequence, for any $\varepsilon > 0 \,\, \exists N \in \mathbb{N}$ s.t. for every $n \geq N$ $|a_n - a| < \varepsilon$. My (not so valiant) efforts:
\begin{align*} 
\left| \frac{an+c}{bn+d} - \frac{a}{b} \right| &< \varepsilon \\
\left| \frac{abn + bc - abn - ad}{b^2n+bd} \right| &< \varepsilon \\
\left| \frac{bc-ad}{bn+d} \right| &< \varepsilon \left| b \right| \\
\left| bn+d \right| &> \left| \frac{bc - ad}{\varepsilon b} \right| \\
\left| bn+d \right| &> \left| \frac{c}{\varepsilon} - \frac{ad}{\varepsilon b} \right|
\end{align*}
I'm used to being able to manipulate constants or remove absolute values when solving these kinds of problems, but with general constants I'm not sure what to do next.
 A: You need $b\ne0$ to prove this. When you arrive at
$$
|bn+d|>\left|\frac{bc-ad}{b\varepsilon}\right|
$$
you're almost done, because this is the same as
$$
\left|n+\frac{d}{b}\right|>\left|\frac{bc-ad}{b^2\varepsilon}\right|
$$
which is certainly satisfied for
$$
n>\left|\frac{bc-ad}{b^2\varepsilon}\right|-\frac{d}{b}
$$
If $b=0$ (and $d\ne0$, then the limit is $\infty$ or $-\infty$ according to $a/d>0$ or $a/d<0$. If also $a=0$, of course, the sequence is constant.
A: Because
$$|bn+d|\ge|bn|-|d|$$
You can require that
$$|b|n-|d|\ge \frac{1}{\epsilon}\left|c-\frac{ad}{b}\right|$$
$$n\ge \frac{1}{|b|\epsilon}\left|c-\frac{ad}{b}\right|+\frac{|d|}{|b|}$$
A: You can do as follows $$M=\left| \frac { bc-ad }{ { b }^{ 2 } }  \right| \\ \left| \frac { bc-ad }{ b\left( bn+d \right)  }  \right| =\left| \frac { bc-ad }{ { b }^{ 2 }\left( n+\frac { d }{ b }  \right)  }  \right| =\frac { M }{ \left| n+\frac { d }{ b }  \right|  } <\frac { M }{ n } <\varepsilon \\ n>\frac { M }{ \varepsilon  } \\$$

$$ { N }_{ \varepsilon  }=\left\lfloor \frac { M }{ \varepsilon  }  \right\rfloor $$

A: 1.Replace n by n=1/t. 
2.Limit becomes t tending to zero.
3.Take 1/t common & cancel.
4.Put t=0.
