Prove that the limit exists:$$\lim\limits_{x \to -3} \frac{x+5}{x+2} = -2$$
My answer is:
Let $\epsilon > 0$, there exists a $\delta=\boxed{min\{\frac{\epsilon}{6},\frac{1}{2}\}}$. Let $x$ such that $0<|x + 3| < \delta$,
$$|f(x) +2| = |\frac{x+5}{x+2} +2| = |\frac{3x+9}{x+2}| = 3 \cdot |\frac{x+3}{x+2}| < 3 \cdot \delta \cdot |\frac{1}{x+2}|$$
Now I need somehow to get rid of that $|\frac{1}{x+2}|$ so I could be able to compare the rightmost value to $\epsilon$ in order for me to show that the limit exists.
So I do the following: $$-\delta < x + 3 < \delta \implies -1-\delta<x+2<\delta-1 \implies \frac{1}{\delta-1} < \frac{1}{x+2} < \frac{1}{-1-\delta}$$
I choose to bound delta as the following $\delta \leq \frac{1}{2}$ and I get that $$-2 < \frac{1}{x+2} < -\frac{2}{3} \implies |\frac{1}{x+2}| < 2$$
So now we'll get $$ 3 \cdot \delta \cdot |\frac{1}{x+2}| < 6\cdot\delta = \epsilon \implies \delta=\frac{\epsilon}{6}$$
This concludes my proof. Now, it has been a few months since I last proved a basic limit like that, and I get every step of the way. However, something bugs me about the way I chose my $\delta$. What would happen if I didn't bound $\delta \leq \frac{1}{2}$? Can I just leave the delta as it is, write $|\frac{1}{x+2}|$ as a function of delta and from there compute epsilon? I can see why someone would bound delta, my guess is that it shortens the calculation of epsilon. Less $min$ and $max$, this way the proof stays cleaner. I never saw something like what I suggested and I was wondering if it is possible. Can't find an excuse why something like that wouldn't be applicable.
Suggestions are greatly appreciated.
We can find the optimal value for $\delta$ in this way, let assume
$$ \left|\frac{x+5}{x+2} +2\right|<\epsilon \implies -2-\epsilon<\frac{x+5}{x+2}<-2+\epsilon \implies -2-\epsilon<1+\frac{3}{x+2}<-2+\epsilon \implies -3-\epsilon<\frac{3}{x+2}<-3+\epsilon \implies \frac{-3-\epsilon}{3}<\frac{1}{x+2}<\frac{-3+\epsilon}{3}\\\implies -\frac{3}{3-\epsilon}<x+2<-\frac{3}{3+\epsilon} \\\implies -\frac{3}{3-\epsilon}+1<x+3<-\frac{3}{3+\epsilon}+1 \\\implies \frac{-\epsilon}{3-\epsilon}<x+3<\frac{\epsilon}{3+\epsilon}\implies|x+3|<min\{\frac{\epsilon}{3+\epsilon},\frac{\epsilon}{3-\epsilon}\}=\frac{\epsilon}{3+\epsilon}=\delta$$