Convergence of Inverse of Convergent Sequence

Solution 1:

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.

Solution 2:

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

$$\;n>K\;,\;\;|x_n-x|<|x|\epsilon M\;\;\implies$$

$$\implies\;\left|\frac1{x_n}-\frac1x\right|=\left|\frac{x-x_n}{xx_n}\right|<\frac{|x|\epsilon M}{|x|M}=\epsilon$$

Solution 3:

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.