We know that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n – 2 \rvert < \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' > 0$, there exists an $N' \in \mathbb{N}$ such that $\left| \frac{1}{x_n} – \frac{1}{2} \right| < \epsilon'$ for all $n \geq N'$.
Let $\varepsilon = \varepsilon' – \frac{3}{2}$. Then by the triangle inequality, we have $$\left| \frac{1}{x_n} – \frac{1}{2} \right| \leq \left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right|.$$
Because we proved earlier that $x_n > 1$ for all $n \geq N$, we know that $$\left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right| < \left| x_n\right| + \left| -\frac{1}{2}\right|.$$
By the triangle inequality again, we know that $$\left| x_n – \frac{1}{2}\right| \leq \left| x_n\right| + \left| -\frac{1}{2}\right|.$$
Note that $\left| x_n – \frac{1}{2}\right| = \left| (x_n – 2) + (2 – \frac{1}{2})\right|$, so we get $$\left| (x_n – 2) + (2 – \frac{1}{2})\right| \leq \left| x_n – 2\right| + \left| 2 – \frac{1}{2}\right|.$$
Then we have $$\left| x_n – 2\right| + \left|2 – \frac{1}{2}\right| < \varepsilon + \frac{3}{2} = \left(\varepsilon' – \frac{3}{2}\right) + \frac{3}{2} = \varepsilon',$$ so $\frac{1}{x_n} \rightarrow \frac{1}{2}$.
I just realized that I don't know how to make sure that $\varepsilon' – \frac{3}{2}$ is a positive number. How can I do this?
Best Answer
In fact there's no reason why $\varepsilon' - \frac{3}{2}$ should be positive.
Instead notice that for all natural numbers $n$ such that $x_n\neq 0$, it holds that $$\left|\dfrac{1}{x_n}-\dfrac1 2\right|=\left|\dfrac{x_n-2}{2x_n}\right|.$$
Relating the RHS to the convergence of $(x_n)_{n\in \mathbb N}$ and to $(x_n)_{n\in \mathbb N}$ being greater (eventually) than $1$ should get you there.