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.

Yes that's right. From this step on, as $x\to -1$ you can write$$-\epsilon<x+1<\epsilon$$ for some  small choice of $\epsilon>0$ therefore $$1-2\epsilon<2x+3<1+2\epsilon$$since for small enough $\epsilon>0$ we have $1-2\epsilon >{1\over 2}$ we can conclude that $$0\le {|1+x|\over |2x+3|}<{\epsilon \over {1\over 2}}=2\epsilon$$which completes the proof.
