epsilon-delta proofs: can delta be defined in terms of both epsilon and x?

Question

I'm trying to construct a proof for the following limit:

$\lim_{x \to -2} \frac{2x+5}{x+4} = \frac{1}{2} $

I know I need to show that for $ \epsilon > 0$ , $\exists \delta > 0$ such that if $0 < |x-(-2)| < \delta$, $\left|\frac{2x+5}{x+4} - \frac{1}{2}\right| < \epsilon$

Here's the steps I took that led me to my question:

$\left|\frac{2x+5 - \frac{1}{2}(x+4)}{x+4}\right| < \epsilon$

$\left|\frac{2x+5 - \frac{1}{2}x - 2}{x+4}\right| < \epsilon$

$\left|\frac{\frac{3}{2}x +3}{x+4}\right| < \epsilon$

$\frac{3}{2} * \left|\frac{x+2}{x+4}\right| < \epsilon$

I know I want to get this value in terms of just $x+2$ so that I can pick my delta value in terms of epsilon, since the limit is defined as $-2$. Since I have that $x+2$ sitting right in the denominator (and this professor is not sadistic so I don't think he would try to make this overly difficult), I'm wondering if it's "legal" for me to pull the whole $|x+4|$ out of the denominator, and do something like:

$\frac{3}{2(x+4)} * \left|x+2\right| < \epsilon$

and then make $\delta = \frac{\epsilon}{\frac{3}{2x+8}} = \frac{\epsilon(2x+8)}{3} $

If I'm not allowed to do this, does anyone have suggestions about where I might go with it next? I'm really not sure how else to get rid of the denominator. Thanks very much!

Answer 1

So, consider the following:

Take a $\delta_{1} = 1: |x+2|<\delta_{1}$, then this conveys that $1<|x+4|$. Then $\dfrac{|x+2|}{|x+4|}<|x+2|$

So it is proved for: $\delta = min(1, \frac{2}{3}\epsilon)$

Answer 2

Let's start with where you left. The inequality which needs to be ensured is $$\frac{|x+2|}{|x+4|}<\frac{2\epsilon}{3}\tag{1}$$ To proceed further we need to bound the denominator from below. Thus we can see that $$|x+4|=|x+2-(-2)|\geq 2-|x+2|$$ If we assume $|x+2|<1$ then we have $|x+4|>1$ and therefore $$\frac{|x+2|}{|x+4|}<|x+2|$$ Thus our target inequality $(1)$ can be ensured if we have $|x+2|<2\epsilon /3$ and $|x+2|<1$. It follows that a suitable value of $\delta$ is $$\delta=\min\left(1,\frac{2\epsilon}{3}\right)$$