Show with the epsilon-delta definition that $\lim_{x \to 2} \frac{1}{x - 1} = 1$ I have an assignment about epsilon-delta proofs and I'm having trouble with this one. I have worked it through using some of the methods I've picked up for similar proofs but it's just something about this particular expression that doesn't sit right with me. Any feedback would be very helpful. This is how far I've come:
Let $\varepsilon > 0$. We want to find a $\delta$ so that $\left|\frac{1}{x - 1} - 1\right| < \varepsilon$ when $0 < |x - 2| < \delta$. We expand the expression:
\begin{align*}
    \left|\frac{1}{x - 1} - 1\right| &< \varepsilon \\
    \left|\frac{1}{x - 1} - \frac{x - 1}{x - 1}\right| &< \varepsilon \\
    \left|\frac{2 - x}{x - 1}\right| &< \varepsilon \\
    |{x - 1}| &< \frac{|x - 2|}{\varepsilon} \\
\end{align*}
We could let $\delta = \dfrac{|x - 2|}{\varepsilon}$ but $|x - 2|$ contains an unwanted variable. Since the limit is only relevant when $x$ is close to $a$ we'll restrict $x$ so that it's at most $1$ from $a$ or in other words, in our case, that $|x - 1| < 1$. This means $0 < x < 2$ and that $-2 < x - 2 < 0$. Looking at our previous inequality
\begin{align*}
    |{x - 1}| &< \frac{|x - 2|}{\varepsilon}
\end{align*}
we see that the right-hand side is the smallest when $|x - 2|$ is the smallest which by the range above is when $x - 2 = -2$ and then we have that
\begin{align*}
    |{x - 1}| &< \frac{|x - 2|}{\varepsilon} < \frac{2}{\varepsilon}
\end{align*}
We now have the two inequalities $|x - 1| < 1$ and $|x - 1| < \frac{2}{\varepsilon}$. Let $\delta = \textrm{min}(1, \frac{2}{\varepsilon})$ and by definition we have that for every $\varepsilon > 0$ there is a $\delta$ so that $|f(x) - A| < \varepsilon$ for every $x$ in the domain that satisfies $0 < |x - a| < \delta$. $\blacksquare$

Assuming wlog $\frac32\le x\le \frac52$ we have
$$\left|\frac{2 - x}{x - 1}\right| < \varepsilon \iff  \left|2 - x\right| < \varepsilon \left|x - 1\right|\le \frac 32 \varepsilon $$
then it suffices to take $\delta <\frac 32 \varepsilon $.
