Use $\epsilon$, $\delta$ to prove that $\lim\limits_{x\to\ b} \frac{1}{a+x}$ = $\frac{1}{a+b}$. I've been working on this epsilon delta proof for the longest time now, and I can't quite get it. 
Let $a>0$ and $b>0$. Use $\epsilon$, $\delta$ to prove that $\lim\limits_{x\to\ b} \frac{1}{a+x}$ = $\frac{1}{a+b}$. 
I've found that I need to prove $|x-b|$ < $\epsilon$ $|a+x|$ $|a+b|$ and that I can do this by calling $\delta$ something greater than $|x-b|$, and that I also need to eliminate $|a+x|$ somehow, but that's where I'm struggling. 
The concept of calling $\delta$ the minimum of a few values has been helpful, but I'm not sure how to apply it here.
Thanks for your thoughts.
 A: I think we should find a way to keep $|x-b|$ on the LHS of the inequality.
For a fixed small $\epsilon>0$, we want to find $\delta>0$ such that $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon$ when $|x-b|<\delta$. Note that $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon \Leftrightarrow \frac{|x-b|}{|a+x||a+b|}<\epsilon \Leftrightarrow \frac{|a+x||a+b|}{|x-b|}>\frac{1}{\epsilon}$.
Moreover, note that $|a+x|=|x-b+b+a|\leqslant |x-b|+|b+a|.$ Thus, it is sufficient to find $\delta>0$ such that $\frac{(|x-b|+|a+b|)|a+b|}{|x-b|}>\frac{1}{\epsilon}$, which is equivalent to $\frac{|a+b|^2}{|x-b|}>\frac{1}{\epsilon}-|a+b|$. And since $\epsilon>0$ is small, we can assume $\epsilon<\frac{1}{|a+b|}$. Then it is equivalent to $|x-b|<\frac{|a+b|^2}{\frac{1}{\epsilon}-|a+b|}$, thus it is sufficient to take $\delta=\epsilon|a+b|^2$ (since $\epsilon|a+b|^2<\frac{|a+b|^2}{\frac{1}{\epsilon}-|a+b|}$).
In $\epsilon-\delta$ language, we can state it as: for each fixed $0<\epsilon<\frac{1}{|a+b|}$, we have $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon$ for all $x$ such that $|x-b|<\epsilon|a+b|^2$.
A: $\forall \epsilon>0, \exists \delta>0$ such that $|x - b|< \delta \implies |\frac 1{a+x} - \frac 1{a+b}| < \epsilon$
$|\frac 1{a+x} - \frac 1{a+b}| = |\frac {b-x}{(a+x)(a+b)}| < \frac {\delta}{|(a+x)(a+b)|}$
If you can prove that $|(a+x)(a +b)|> 0$ then you are done.  (which is pretty much what you told us, I just need to do certain things for myself)
$|a + x| = |a+b + (x-b)| \ge |a+b| - |x-b| \ge |a+b| - \delta$
let $\delta \le \min ( \frac {|a+b|}{2}, \epsilon M)$
In which case:
$|a + x| \ge \frac {|a+b|}{2}\\
|a + x||a+b| \ge \frac {(a + b)^2}2$
$|x-b|<\delta \le \min (\frac {(a + b)}2,\frac {(a + b)^2}2 \epsilon) \implies |\frac 1{a+x} - \frac 1{a+b}| < \epsilon$
A: Let $\varepsilon > 0$. Want some $\delta > 0$ such that $0 < |x-b| < \delta$ implies $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$. We start with the inequality $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$ and try to manipulate it to give us insight into what $\delta$ should be. We have
\begin{align}
\frac{1}{a + x} - \frac{1}{a + b} &= \frac{a+b - (a+x)}{(a+x)(a+b)} \\
&= \frac{1}{b+a} \cdot \frac{b-x}{a+x} \\
\implies \left| \frac{1}{a + x} - \frac{1}{a + b} \right| &= \frac{1}{b+a} \cdot \frac{\left| b-x \right|}{\left| a+x \right|}. \tag{1}
\end{align}
We want to establish an upper bound for the LHS, which is accomplished by picking the largest possible RHS. This is found by maximizing $|b - x|$ and minimizing $|a - x|$. Our selection of $\delta$ determines which values of $x$ are allowed:
$$
0 < |x - b| < \delta \iff x \in (b-\delta, b) \cup (b, b+\delta). \tag{2}
$$
Certainly $(2)$ implies the maximum of $|b - x|$ is $\delta$. The minimum of $a+x$ is $a + b - \delta$, and hence 
$$
\text{minimum of } |a + x| = \begin{cases}
a + b - \delta & a + b - \delta > 0 \\
0 & \text{otherwise}
\end{cases}. \tag{3}
$$
Armed with this info, let's try to pick a reasonable $\delta$ to bound the LHS of $(1)$. By $(3)$ we had better 
$$
\text{enforce $\quad \delta < a + b$}. \tag{4}
$$
If we do, by $(1)$ we have
\begin{align*}
\left| \frac{1}{a + x} - \frac{1}{a + b} \right| & \leq \frac{1}{b+a} \cdot \frac{\delta}{a + b + \delta} \\
& \leq \frac{1}{b+a} \cdot \frac{\delta}{a + b}. \tag{5}
\end{align*}
Now, if we also
$$
\text{enforce     $\quad \delta < \varepsilon \cdot (a + b)^2$}, \tag{6}
$$
then $(5)$ becomes
$$
\left| \frac{1}{a + x} - \frac{1}{a + b} \right| \leq \frac{1}{b+a} \cdot \frac{\delta}{a + b} < \frac{1}{(a + b)^2} \cdot \varepsilon \cdot (a + b)^2 = \varepsilon.
$$
That is, if we enforce $(4)$ and $(6)$, i.e., $\delta < \min\{a + b, \varepsilon \cdot (a+b)^2\}$, then $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$, as desired. 
A: The expression you want to keep small is
$$
\frac{1}{a+x}-\frac{1}{a+b}
=\frac{b-x}{(a+b)(a+x)}
$$
and you can assume $x>0$ by taking $|x-b|<b$. Then $a+x>a$ and so $1/(a+x)<1/a$.
Therefore
$$
\left|\frac{1}{a+x}-\frac{1}{a+b}\right|=
\left|\frac{b-x}{(a+b)(a+x)}\right|<
|b-x|\frac{1}{a(a+b)}
$$
So, if
$$
|b-x|<\varepsilon a(a+b)
$$
you have the required inequality. So you can take
$$
\delta=\min(b,\varepsilon a(a+b))
$$
