Proof Verification of $\lim _{x\to 3}\left(\frac{2}{x+3}\right)=\frac{1}{3}$ 
Using the epsilon delta definition to prove: $f$ is continuous at $a$ if for every $  ε > 0$ there exists $δ > 0$ such that if $0 < |x - a| < δ$ then $| f (x) - f(a)| < ε.$

$$\lim _{x\to 3}\left(\frac{2}{x+3}\right)=\frac{1}{3}$$
$$0 < |x - 3| < δ\,\text{ then }\,\left| \frac{2}{x+3} - \frac{1}{3}\right| < ε.$$
After some simplification I get:
$$\frac{|x-3|}{|3||x+3|} < ε$$
$|x-3|$ is bounded by $δ $, $|3|$ is a constant, so $|x+3|$ is the only thing that I have worry about. I have to control it so it doesn't become $0$ or else I got some problems.
Since for every $  ε > 0$ there exists $δ > 0$, I will pick a $δ > 0$. So let $δ = 1$
$$0 < |x - 3| < δ$$
$$-1 < x - 3 < 1$$
plus 6 to both side to "induce" $x+3$, I get $5 < x - 3 < 7$, now this is nicely bounded, so using this
I know that $\frac{|x-3|}{|3||x+3|} < \frac{δ}{3 \times 5}$
Now this is where my problem is
Can I just say  $\frac{|x-3|}{|3||x+3|} < \frac{δ}{3 \times 5} < \epsilon$?
If I am allowed to, why? I thought I have prove until I get $\frac{|x-3|}{|3||x+3|} < ε$ ? If I were to do this, didn't I just assumed that $\frac{δ}{3 \times 5} < \epsilon$?
Can someone please continue from here? If I take the min{} of the deltas can I conclude straight away that $| \frac{2}{x+3} - \frac{1}{3}| < ε.$ If this is the case, why?
 A: For any $\epsilon\gt 0$, you can't say that $\frac{|x-3|}{|3||x+3|} < \frac{δ}{3 \times 5} < \epsilon$ unless you have shown that corresponding to every $\epsilon\gt 0,$ there exists a $\delta \gt 0$ such that the limit definition conditions are met.   
We can create such $\delta$, mutatis mutandis, as below also:
Note that if $\delta=1$ such that $|x-3|\lt 1$, then it follows that $2\lt x\lt 4\implies 5\lt x+3\lt7\implies \frac 17\lt \frac 1{x+3}\lt \frac 15$. Now, we have
$\left| \frac{2}{x+3} - \frac{1}{3}\right| =\frac{|x-3|}{|3||x+3|} \lt \frac 1{15}|x-3|$
Given any $\epsilon\gt 0$, choose $\delta'= \min\{1, 15\epsilon\}$ 
Now, we have for any $\epsilon\gt 0, \exists \delta'=\min\{1,15\epsilon\}\gt 0: |x-3|\lt \delta' \implies \left| \frac{2}{x+3} - \frac{1}{3}\right| \lt \epsilon$.
A: This is a common problem faced by many students when decoding $\epsilon, \delta$ definition of limit.
Here is a simple problem to ponder over. Let $a>0$ and you are supposed to prove that $a+(1/a)>3/2$. Now assume that you are somehow able to prove that $a+(1/a)\geq 2$. Do you see that this is sufficient to solve your original problem simply because $2>3/2$? If you have any doubts here please go through it once again.

Next we come to current question. Here our goal is to establish $$\left|\frac{2}{x+3}-\frac{1}{3}\right|<\epsilon \tag{1}$$ Now this is a slightly more complicated problem that the one I discussed earlier about $a+(1/a)$ precisely because of two reasons:

*

*$\epsilon$ is not a specific number like $1,2$ but rather is an arbitrary positive number. This means that the inequality has to be established for all such values of $\epsilon$ and not for some specific value like $1$.

*the means to establish this goal are limited. We have to constrain the values of $x$ in a very specific manner in order to establish $(1)$. The specific manner is choosing  a suitable $\delta>0$ and constraining $x$ by $0<|x-3|<\delta$.

The first problem described above is simple. We deal with $\epsilon $ symbolically and do not assume anything beyond $\epsilon>0$. The second problem is the crux of the definition of limit.
The way out is to simplify our goal $(1)$ so that it can be achieved by the specific kind of constraints on $x$. We first start by algebraic manipulation and the goal $(1)$ is equivalent to $$\frac{|x-3|}{|x+3|}<3\epsilon \tag{2}$$ In the example we discussed at the beginning of the answer we replaced the goal $a+(1/a)>3/2$ with another (but not equivalent) goal $a+(1/a)\geq 2$. We can use the same idea of replacing $(2)$ with something which is not necessarily equivalent but is sufficient to ensure $(2)$.
The fact that we can replace our original target with something simpler but not equivalent is a big saving grace and makes our life considerably simpler. In practice we replace $(2)$ with two further sub-goals
\begin{align}
\frac{|x-3|}{|x+3|} &< g(x)\tag{3a}\\
g(x) &< 3\epsilon\tag{3b}
\end{align}
And this need not be equivalent to $(2)$ but should be sufficient to conclude $(2)$ ie $(3a),(3b)$ together must imply $(2)$. The function $g(x) $ is usually chosen to be very simple in form which makes handling of $(3b)$ almost trivial. In particular a common choice for $g(x) $ is to be of the form $K|x-3|$ where $K$ is some positive constant.
Next we need to choose suitable values of $\delta$, say $\delta_1,\delta_2$ to constrain $x$ such that both $(3a),(3b)$ are achieved respectively. Here we take another advantage based on form of $g(x)=K|x-3| $. We take a suitable $\delta_1 $ almost by our own choice to fulfill $(3a)$ and fix $K$ accordingly (rather than choosing $K$ first and finding $\delta_1$ accordingly). Thus in your question you take $\delta_1=1$ and $K=1/5$. Clearly $\delta_2=15\epsilon $ is sufficient to ensure $(3b)$. And if $$\delta=\min(\delta_1,\delta_2)=\min(1,15\epsilon)$$ then the constraint $0<|x-3|<\delta$  ensures both $0<|x-3|<\delta_1$ and $0<|x-3|<\delta_2$ so that both the targets $(3a),(3b)$ are achieved and together they ensure that target $(2)$ is also achieved.

For many limit problems you may have to replace $|f(x) - L |<\epsilon $ by $$g(x) <\epsilon /2,h(x)<\epsilon /2$$ and each of these two sub-goals may further need to be replaced by goals like $(3a),(3b)$. In general there is huge flexibility in replacing one target with multiple different targets because we just need sufficiency and not equivalence in such replacements.
A: Hint: when $x$ is in neighbourhood of $3$, for example for $\delta=1$,  then $|x+3|>1\Rightarrow \frac{1}{|x+3|} \lt 1$.
