Prove that if $x_n → 2$, then $\frac1{x_n}→\frac12$

Question

Prove that if $x_n → 2$ , then $ \dfrac{1}{x_n} → \dfrac{1}{2}$.

Attempt:

$$ \left|\frac{1}{x_n} - \frac{1}{2} \right| < ε\\ ⇔ -ε < \frac{1}{x_n} - \frac{1}{2} < ε\\ ⇔ -ε < \frac{2-x_n}{2x_n} < ε\\ ⇔ ε > \frac{x_n -2}{2x_n} > -ε $$

Answer 1

1)Let $\epsilon \gt 0$ be given.

There is a $n_0$ such that for $n \ge n_0$

$|x_n - 2| < \epsilon$.

2) Consider $\epsilon =1$.

There is a $n_1$ such that for $n \ge n_1$

$|x_n-2| \lt 1$, or $1<x_n<3$.

3) $|1/x_n -1/2| = \dfrac{|x_n - 2|}{|2x_n|}\lt$

$\dfrac{|x_n-2|}{2} \lt |x_n-2|$.

For $\epsilon \gt 0$ given (cf.1), and

$n_2 \gt \max(n_0,n_1)$ we have :

$|1/x_n-1/2| \lt |x_n -2| \lt \epsilon$.

Answer 2

When you claim $\frac 1{x_n} \to \frac 12$, if somebody gives you an $\epsilon \gt 0$ you have to be able to respond with an $N$ so that if $n \gt N, \left|\frac 1{x_n} - \frac 12 \right| \le \epsilon$. You have been told $x_n \to 2$ by somebody, so you can give them an $\epsilon' \gt 0$ and get back an $N'$. You will use their $N'$ as your $N$ (though you could use anything higher), you just need to figure out what $\epsilon'$ to give them based on the $\epsilon$ you received.

At the end of your question you have $$\epsilon \gt \frac {x_n-2}{2x_n} \gt -\epsilon$$ As $\epsilon$ is supposed to be small, it is tempting just to plug in $2$ for the $x_n$ in the denominator, which would say $4\epsilon \gt x_n-2 \gt -4\epsilon$ and you could take $\epsilon'$ in the above to be $4\epsilon$. This is a bit too aggressive because $x_n$ could be smaller than $2$. It is better to say you will never use an $\epsilon'$ greater than $\frac 12$ so you know $x_n \gt \frac 32$ and then you get $3\epsilon \gt x_n-2 \gt -3\epsilon$ You can then ask for the $N$ that is sufficient to guarantee $|x_n-2| \lt \min(\frac 12,3\epsilon)$ and use that to guarantee $\left|\frac 1{x_n}-\frac 12\right| \lt \epsilon$ if $n \gt N$