TITLE: Convergence of Inverse of Convergent Sequence
QUESTION [5 upvotes]: Let $\{x_n\}$ be a sequence in $\mathbb{R}$ where $\forall n\in\mathbb{N}:x_n\neq 0$ and it converges to some $x\neq 0$. If the sequence is NOT monotone is it ever true that $\frac{1}{x_n}\rightarrow\frac{1}{x}$? If so what other conditions (if any) are needed and how would you show it.  
Thanks in advance any feedback is greatly appreciated.

REPLY [3 votes]: Suppose $\;x_n\to x\neq 0\;$ , and $\;x_n\neq 0\;$ (enough to assume for almost all $\;n\in\Bbb N\;$ and then throw away all the zero terms). 
Now there exists $\;M\in\Bbb R^+\;$ s.t. $\;|x_n|\ge M\;\;\forall\,n\in\Bbb N\;$ (why?), and then for all $\;\epsilon >0\;$ there exists $\;K\in\Bbb N\;$ s.t. that for all 
$$\;n>K\;,\;\;|x_n-x|<|x|\epsilon M\;\;\implies$$
$$\implies\;\left|\frac1{x_n}-\frac1x\right|=\left|\frac{x-x_n}{xx_n}\right|<\frac{|x|\epsilon M}{|x|M}=\epsilon$$