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.

The neighborhood you have chosen is too large, it contains a root of the denominator. Let us try with the bounds
$$-1\pm\frac13$$ (any deviation smaller than $\dfrac12$ can do).
The function is monotonous in that range and the extreme values of $\left|\dfrac7{2x+3}\right|$ are
$$21, \frac{21}{5}$$ so you can write
$$\left|\frac{x+5}{3x+2}-4\right|<21|x+1|.$$
