Convergence of Inverse of Convergent Sequence 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.
 A: It is always true, provided that $x\neq 0$ and each $x_n\neq 0$.
Generally, $f(x_n)\to f(x)$ if $f$ is continuous and $x$ and each $x_n$ are in the domain of $f$. This follows from continuity.
A: Suppose $x{}_n\to x\neq 0$ and $x{}_n\neq 0$ (enough to assume $x{}_n\to x\neq 0$, then $x{}_n\neq 0$ for almost all $n\in\Bbb N$ - i.e. all but finitely many - and we can throw away all the zero terms).
Now there exists $M>0$ s.t. $|x_n|\ge M$ for all $n\in\Bbb N$ (see the addendum), and then for all $\epsilon >0$ there exists $K\in\Bbb N$ s.t. that for all $n>K$
$$|x_n-x|<|x|\epsilon M\implies\left|\frac1{x_n}-\frac1x\right|=\left|\frac{x-x_n}{xx_n}\right|<\frac{|x|\epsilon M}{|x|M}=\epsilon.$$
Addendum: Since $\lim_{n\to\infty}|x_n|=|\lim_{n\to\infty} x_n|=|x|>0$ we can choose some $m$ such that $0<m<|x|$ and by the definition of convergence there is some $N\in\mathbb N$ such that $m\leq |x_n|$ for all $n> N$. Now consider $M:=\min\{|x_1|,\ldots,|x_N|,m\}$.
A: Consider the mapping: $g: \mathbb{R}^+\to \mathbb{R}^+, \quad x \mapsto \frac{1}{x}$. $g$ is continuous in its domain. This implies that if $x_n \to x$ then $g(x_n) \to g(x)$, that is your thesis.
Same argument can be used in the case of negative values.
EDIT:
As MPW wrote it is sufficient to take $g: \mathbb{R} \setminus \{0\}\to \mathbb{R} \setminus \{0\}$ without splitting cases.
