Use the definition of limit to show that $\lim_{x \to -1} \frac{x+5}{2x+3}=4$ 
Use the definition of limit to show that
$\lim_{x \to -1} \frac{x+5}{2x+3}=4$

We have $|\frac{x+ 5}{2x+3} -4|= |\frac{x+5-8x-12}{2x+3}|=|\frac{-7x-7}{2x+3}|=\frac{7}{|2x+3|}|x+1|$
To get a bound on the coefficient of $|x+1|$, we restrict $x$ by the condition $-2<x<0$ [neighborhood of $-1$] . For $x$ in this interval, we have $-1<2x+3<3$, so that
$|\frac{x+ 5}{2x+3} -4|= \frac{7}{|2x+3|}|x+1|<...$
Is that true, please? And I don’t know how can I complete.

You way of thinking is on the right track. First, we need to "guess" $\delta$ so we will make a draft of the calculation first.
This is just the "draft" :
\begin{align*}
\big|\frac{x+5}{2x+4}-4\big| &< \varepsilon \\
7\big|\frac{x+1}{2x+3}|&< \varepsilon \\
|x+1| &< \frac{\varepsilon}{7}|2x+3|
\end{align*}
So now, for convenience, I bound $|x+1|<\frac{1}{4}$. You can bound with any convenient number. Observe that
\begin{align*}
-\frac{1}{4} &< x + 1 < \frac{1}{4} \\
-\frac{1}{2} &< 2x + 2 < \frac{1}{2} \\
\frac{1}{2} &< 2x + 3 < \frac{3}{2}
\end{align*}
Now, we are all set. From the calculation above, we can fix $\varepsilon>0$ and set $\delta = \min\{\frac{\varepsilon}{14},\frac{1}{4}\}$
Thus, we will obtain :
\begin{align*}
|x+1| &< \frac{\varepsilon}{14}\\
|x+1| &< \frac{\varepsilon}{7} . \frac{1}{2}\\
|x+1| &< \frac{\varepsilon}{7}|2x+3|\\
|7x+7| &< \varepsilon|2x+3|\\
\big|\frac{x+5}{2x+3}-4\big| &< \varepsilon
\end{align*}
Thus, we have completed the proof.
