Prove that $\frac{1}{x_n}\to\frac{1}{2}$ if $x_n\to 2$. No limit theorems allowed. Thanks!
We know it is something like 
$$\left|\frac{1}{x_n}-\frac{1}{2}\right|=\left|\frac{2-x_n}{2x_n}\right|<\epsilon$$
 A: Hint Since $x_n \to 2$, there exists some $N$ so that for all $n>N$ you have $x_n >1$.
Then, for all $n>N$ you have:
$$\left|\frac{1}{x_n}-\frac{1}{2}\right|=\left|\frac{2-x_n}{2x_n}\right|<\left|\frac{2-x_n}{2}\right|$$
A: Let $\epsilon>0$. Since we care only about what happens when $\epsilon$ is small, we can also assume that $\epsilon<1$. (You’ll see in a bit why I want this.) Since $x_n\to 2$, there is an $m\in\Bbb N$ such that $|x_n-2|<\epsilon$ whenever $n\ge m$. Thus, for $n\ge m$ we have
$$\left|\frac1{x_n}-\frac12\right|=\left|\frac{2-x_n}{2x_n}\right|=\frac{|x_n-2|}{2|x_n|}<\frac{\epsilon}2\;,$$
because $|x_n-2|<\epsilon$ and $|x_n|>1$ (why?).
A: Continuing where you left off:
$$
|\frac{2-x_n}{2x_n}| = |\frac{2-x_n}{x_n - 2 + x_n - 2 + 4}| \leq \frac{|2-x_n|}{|x_n - 2| + |x_n - 2| + 4}
$$
Since  $x_n\to 2$, for any $\delta$ there exists an $N$ such that if $n>N, |2-x_n| < \delta$. Thus we need only find a $\delta(\epsilon)$ such that for a given $\epsilon > 0$, 
$$
\frac{\delta}{2\delta + 4} \leq \epsilon
$$
Hint: There's an easy simplification you can make to the above which makes finding the $\delta$ very simple.
