<s>[INST] Prove $\lim_{x\to 3}\frac{1}{x+1}=\frac{1}{4}$ using $\epsilon$-$\delta$ definition of a limit First of all, I'd just like to say I'm fairly new to proving limits using $\epsilon-\delta$ definition, so I apologize in advance if I ask a very obvious question, or make an elementary error.

Given $\displaystyle\lim_{x\to 3}\frac{1}{x+1}=\frac{1}{4}$, I want to prove it using the $\epsilon-\delta$ definition of a limit: 
The definition: $\forall\epsilon>0, \exists\delta>0\ni0<|x-c|<\delta\implies|f(x)-L|<\epsilon$
The definition applied to my example: $\forall\epsilon >0, \exists\delta >0\ni0<|x-3|<\delta\implies\left|\dfrac{1}{x+1}-\dfrac{1}{4}\right|<\epsilon$.

$\text{Scratch Work For Determining}\space \delta$:
$$\left|\dfrac{1}{x+1}-\dfrac{1}{4}\right|=\left|\dfrac{4-(x+1)}{(x+1)\cdot4}\right|=\left|\dfrac{4-x-1}{(x+1)\cdot 4}\right |=\left|\dfrac{3-x}{(x+1)\cdot 4}\right|=\dfrac{|x-3|}{|4(x+1)|}<\epsilon$$
I can let the latter inequality be $|x-3|<\epsilon\cdot|4(x+1)|=4\epsilon\cdot|(x+1)|$, and then try to bound it:
Let $\delta=1$:
$$|x-3|<\delta=1\implies -1<x-3<1\implies 3<x+1<5$$. Plugging the value $5$ instead of $|(x+1)|$, gives the following:
$$|x-3|<4\epsilon\cdot 5=20\epsilon.$$ 
At this stage, I'd then say $\delta=\min\{1,20\epsilon\}$. 
I was told this is not the correct answer though. Can anyone explain why it's not right?
 [/INST] To maximize a fraction, we want to minimize its denominator. With this in mind, given any $\epsilon>0$, let $\delta = \min\{1,12\epsilon\}$. Then if $0 < | x-3| <\delta$, notice that:
\begin{align*}
0<|x-3|<\delta \leq 1
&\implies -1 < x-3 < 1 \\
&\implies 3 < x+1 < 5 \\
&\implies 12 < 4(x+1) < 20 \\
&\implies  \frac{1}{12} > \frac{1}{4(x+1)} > \frac{1}{20} \\
&\implies  \frac{1}{20} < \frac{1}{4(x+1)} < \frac{1}{12} \\
&\implies  \frac{-1}{12} < \frac{1}{20} < \frac{1}{4(x+1)} < \frac{1}{12} \\
&\implies  \left| \frac{1}{4(x+1)} \right| < \frac{1}{12} \\
\end{align*}
Hence, it follows that:
\begin{align*}
\left| \frac{1}{x+1} - \frac{1}{4} \right|
&= \left|\frac{1}{4(x+1)}\right||x-3| &\text{using your work}\\
&< \frac{1}{12}|x-3| & \text{from above}\\
&< \frac{1}{12}(12\epsilon) & \text{since $|x-3|<\delta \leq 12\epsilon$}\\
&= \epsilon
\end{align*}
as desired.
</s>