$\epsilon-\delta$ proof of $\lim_{ x\to 5} \frac{1}{x-4}=1$ $$\lim_{x\rightarrow 5} \frac{1}{x-4}=1$$
So far I have started the proof
For every $\epsilon > 0$ there exists a $\delta>0$ such that $\left|\frac{1}{x-4}-1\right|< \epsilon$, whenever $0 < |x-5| < \delta.$
I am having trouble figuring out what I need to factor out and getting to that step. 
 A: We see $$\left \lvert \frac{1}{x-4} - 1\right \rvert = \left \lvert \frac{1}{x-4} - \frac{x-4}{x-4}\right \rvert = \left \lvert \frac{5-x}{x-4} \right \rvert.$$ For fixed $\epsilon > 0$, take $\delta = \min\{1/2, \epsilon / 2  \}$. Then for $\lvert x - 5 \rvert < \delta$ we have $\lvert x - 4 \rvert > 1/2$ so $\frac{1}{\lvert x -4 \rvert} < 2$. Thus $$\left \lvert \frac{1}{x-4} - 1\right \rvert =\frac{1}{\lvert x-4 \rvert} \lvert x - 5 \rvert < 2\delta < \epsilon.$$
A: Note that:
$$\Big|\frac{1}{x-4} - 1\Big| = \Big|\frac{1}{x-4} - \frac{x-4}{x-4}\Big| = \Big|\frac{5-x}{x-4}\Big| = \Big|\frac{x-5}{x-4}\Big|$$
(The last equality follows since $|c| = |-c|$ for all $c \in \mathbb{R}$.)
Anyway: The final form of the absolute value now has a numerator that you can control. But what will you do about the denominator?
One idea is to pick $\delta$ as being at least as small as something; e.g., if you know that $\delta = \min\{1/2, s\}$ where $s$ is ... something to be determined, then how far can $x$ be from $4$? Answering this question (and recalling that $|x-4|$ can be interpreted as the distance from $x$ to $4$) can allow you to control the entire expression for a well-chosen $s$ expressed in terms of $\varepsilon$.
(For a fully worked-out version, see User8128's answer.)
A: so far so good.
$\left|\frac{1}{x-4}-1\right|\\
\left|\frac{1 - x + 4}{x-4}\right|\\
\left|\frac{x -5}{x-4}\right|\\
$
let $\delta$ be $< \frac12\\
 |x -5| < \delta\\
 |x -4| > \frac 12$
$\left|\frac{x -5}{x-4}\right|< 2\delta$
All of the above is technically scratch work and doesn't need to appear in the proof.
At this point we go back to the beginning.
$\forall \epsilon>0,  \delta = \min(\frac12, \frac12\epsilon), |x-5|<\delta \implies |\frac1{x-4} - 1|< \epsilon.$ 
