$\epsilon-\delta$ 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. 

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 $$\left \lvert \frac{1}{x-4} - 1\right \rvert =\frac{1}{\lvert x-4 \rvert} \lvert x - 5 \rvert < 2\delta < \epsilon.$$
