# Proving a limit through the delta-epsillon definition of a limit, discarding the delta upper bound

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.

• You should claim "$\forall \epsilon > 0$ there exists a $\delta>0$ such that..." – gimusi Jan 27 '18 at 19:21
• @gimusi What you are citing is the definition itself, however whilst proving a limit one should start with "Let $\epsilon > 0$ there exists $\delta$..." and show that there indeed is a delta that suits every epsilon. – 0rka Jan 27 '18 at 19:23
• Maybe it is only a detail but if you are proving at the beginig you don't know the correct value for $\delta$. You should set $|f(x)+2|< \epsilon$ and then find $\delta$ such that... – gimusi Jan 27 '18 at 19:30
• @gimusi You are totally right, that is why the value of $\delta$ is boxed. Indeed while writing the proof you do not know the value of $\delta$, but because I already proved it and I do know the value in advance - I've put it there to be more consistent. – 0rka Jan 27 '18 at 19:33
• Orka.Sorry missed your point. – Peter Szilas Jan 27 '18 at 21:39

## 1 Answer

We can find the optimal value for $\delta$ in this way, let assume

$$\left|\frac{x+5}{x+2} +2\right|<\epsilon \implies -2-\epsilon<\frac{x+5}{x+2}<-2+\epsilon \implies -2-\epsilon<1+\frac{3}{x+2}<-2+\epsilon \implies -3-\epsilon<\frac{3}{x+2}<-3+\epsilon \implies \frac{-3-\epsilon}{3}<\frac{1}{x+2}<\frac{-3+\epsilon}{3}\\\implies -\frac{3}{3-\epsilon}<x+2<-\frac{3}{3+\epsilon} \\\implies -\frac{3}{3-\epsilon}+1<x+3<-\frac{3}{3+\epsilon}+1 \\\implies \frac{-\epsilon}{3-\epsilon}<x+3<\frac{\epsilon}{3+\epsilon}\implies|x+3|<min\{\frac{\epsilon}{3+\epsilon},\frac{\epsilon}{3-\epsilon}\}=\frac{\epsilon}{3+\epsilon}=\delta$$

• Interesting. Indeed this way we don't put an upper bound on $\delta$. Also, in the last line of your answer shouldn't it be $max\{|\frac{-\epsilon}{3-\epsilon}|,|\frac{\epsilon}{3+\epsilon}|\}$ ? – 0rka Jan 27 '18 at 20:14
• @0rka in the inequality $$\frac{-\epsilon}{3-\epsilon}<x+3<\frac{\epsilon}{3+\epsilon}$$ x+3 is between a negative and a positive value thus the condition on |x+3| is satisfied by the minimum of the absolute values of the bounding term. Try with some numerical value to convince yourself that the value for $\delta$ is optimal. – gimusi Jan 27 '18 at 20:37
• What about $\epsilon = 6$? Your inequality will be: $2 < x + 3 < \frac{2}{3}$. – 0rka Jan 27 '18 at 20:40
• I think I get it now. The fact that you would need an upper bound on $\delta$ is because if it wasn't the case then $\frac{1}{\delta -1} < x + 2 < \frac{1}{-\delta-1}$ wouldn't be true for every value of $\delta > 0$, for example $\delta = 6$. Seems right? – 0rka Jan 27 '18 at 20:45
• @0rka Yes ok if you have to fullfil all the real value we need to revise the derivation, since we are dealing with arbitrary "small" values we can bound in order to make sense for the inequalities. In this case we only need that $\epsilon<3$ to derive the optimal value for $\delta$. Otherwise the optimal value becomes the LHS. – gimusi Jan 27 '18 at 20:49