$\epsilon-\delta$ proof of $\lim_{ x\to 5} \frac{1}{x-4}=1$

Question

$$\lim_{x\rightarrow 5} \frac{1}{x-4}=1$$

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.

Answer 1

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

Answer 2

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.)

Answer 3

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.

$\forall \epsilon>0, \delta = \min(\frac12, \frac12\epsilon), |x-5|<\delta \implies |\frac1{x-4} - 1|< \epsilon.$