I'm going crazy trying to understand this problem, like it's literally driving me mad. Please someone help explain this to me?

Use the definition of the limit to show that

$\displaystyle \displaystyle\lim_{x \to -1}(\frac{x+5}{2x+3}) = 4$

So I said "Given any $\displaystyle \epsilon > 0$, there exists a $\displaystyle \delta > 0$ such that if $\displaystyle 0 < \mid x+1 \mid < \delta$, then $\displaystyle \mid \frac{x+5}{2x+3} - 4 \mid < \epsilon$

So I simplified $\displaystyle \mid \frac{x+5}{2x+3} - 4 \mid$ to $\displaystyle 7 \mid \frac{x+1}{2x+3} \mid$, and I get stuck.

The biggest issue I have is that the "answer" (but it really doesn't give you a proof) is in the back of the book, and I'm pulling my hair out trying to understand it.

This is what they say:

If $\displaystyle \mid x+1 \mid < \frac{1}{4}$, then $\displaystyle \mid \frac{x+5}{2x+3}-4 \mid < 7 \mid \frac{x+1}{2x+3} \mid$

Q: How did they say that $\displaystyle \mid x+1 \mid < \frac{1}{4}$. Why were they able to just say that it is less than 1/4?

Then this is where I really get confused:

$\displaystyle 7 \mid\frac{x+1}{2x+3}\mid < 14\mid x+1\mid$

Q: Where did they get this from? I understand that it is less than that, but I don't get how they got that. Like if I were doing a different problem, would I just pull that out of thin air?

Then they say we may take $\displaystyle \delta := inf \{\frac{1}{4}, \frac{\epsilon}{14}\}$

Q: I kind of understand this, so that it will always be less than $\displaystyle \epsilon$, but I'm again not sure where they got this?

If anyone could help me, that would be awesome. I really want so badly to understand how to do this, I just can't do it alone!

Thank you.