<s>[INST] Prove that $\lim_{x\to 0}\frac{x+a}{1-3x}=a$, $a>0$, by $\epsilon$-$\delta$ arguments We have
\begin{align}
\left| \frac{x+a}{1-3x} - a \right| &= \left| \frac{x+a-a+3ax}{1-3x} \right| \\[1em]
&= \left| \frac{x(1+3a)}{1-3x}\right|\\[1em] 
&=  \frac{|x|(1+3a)}{|1-3x|} <\epsilon.
\end{align}
By stating  that $|x|<\delta$, we can proceed by
\begin{align}
 |x| <\frac{\epsilon|1-3x|}{1+3a}.
\end{align}
By restricting $-1/3<x<1/3$ we have that for such $x$ and for  any  $\epsilon$, the corresponding $\delta$ is $\min\{1, 2\epsilon/(1+3a) \}$.  
Is this right? 
 [/INST] This is one of those cases where starting off with $\delta=1$ is not optimal. If you let $\delta=\frac{1}{6}$, then note that
\begin{align}
|x|<\frac{1}{6}&\Longleftrightarrow -\frac{1}{6}<x<\frac{1}{6}\\[1em]
&\Longleftrightarrow \frac{1}{6}>-x>-\frac{1}{6}\\[1em]
&\Longleftrightarrow \frac{1}{2}>-3x>-\frac{1}{2}\\[1em]
&\Longleftrightarrow \frac{3}{2}>1-3x>\frac{1}{2}\\[1em]
&\Longleftrightarrow \frac{2}{3}<\frac{1}{1-3x}<2\\[1em]
&\Longleftrightarrow \frac{1+3a}{1-3x}<2+3a.
\end{align}
Thus, we should choose $\delta=\min\left\{\frac{1}{6},\frac{\epsilon}{2+3a}\right\}$. To see that this choice of $\delta$ works, consider the following:
Given $\epsilon>0$, let $\delta=\min\left\{\frac{1}{6},\frac{\epsilon}{2+3a}\right\}$. If $|x|<\frac{1}{6}$, then $\left|\frac{1+3a}{1-3x}\right|<2+3a$. Also, $|x|<\frac{\epsilon}{2+3a}$. Hence,
$$\left|\frac{x+a}{1-3x}-a\right|=|x|\left|\frac{1+3a}{1-3x}\right|<\frac{\epsilon}{2+3a}\cdot(2+3a)=\epsilon,
$$
as desired.
</s>