I'm having some trouble figuring out how to do this problem. I understand that I have the assumption: for all $\epsilon >0$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$ we have $$|x_n-2|<\epsilon$$ And I know I should be trying to show that $$\left| \frac{1}{x} - \frac{1}{2} \right| < \epsilon.$$ Now I see that $$\left| \frac{1}{x_n} - \frac{1}{2} \right|=\left| \frac{2-x_n}{2x_n} \right| = \frac{|2-x_n|}{|2x_n|}=\frac{|x_n-2|}{|2x_n|}$$ but I'm unsure how to progress from here. I want to go and say something like "$|2x_n|\leq 2M$", where $M$ is the bound of $(x_n)$ since it converges, so we have $$\frac{|x_n-2|}{2M} \leq \frac{|x_n-2|}{|2x_n|}$$ but this doesn't seem to help since the inequality is facing the opposite direction I'd want it to.

Could someone provide a hint on how to progress? Please don't provide any full solutions, thank you.

  • 1
    $\begingroup$ Now, since $x_n\to 2$, there exists $M\in\Bbb N$ such that $|x_n|\geq 1/2$ for all $n\geq M$. Equivalently, $|\frac 1{2x_n}|\leq 1$. Can you take it from here? $\endgroup$ – Prasun Biswas Mar 23 '18 at 6:18
  • $\begingroup$ Please avoid unnecessary mathematical symbols in titles. (+1) for showing your tries. $\endgroup$ – Did Mar 23 '18 at 6:29


As $x_n \to 2$, there is $W>0$ such that if $n >W$ then $|x_n -2|< 1$, that is

$1<x_n<3$ and from there you can get a bound on $\frac1{x_n}$ when $n > W$.

  • $\begingroup$ Slow down: although your post deals with trivialities, it is marred with misprints. $\endgroup$ – Did Mar 23 '18 at 6:28
  • $\begingroup$ thanks Did for pointing out my mistakes. $\endgroup$ – Siong Thye Goh Mar 23 '18 at 6:33

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