**Theorem** The function $f: \mathbb{R} \backslash \{ 0 \} \to \mathbb{R}$ given by $f(x) =\frac{1}{x}$ is [[Continuous Real Function|continuous]] at $c \neq 0$. **Proof (using [[Continuous Function (Epsilon-Delta Definition)|epsilon-delta]])** Fix $c \in \mathbb{R} \backslash \{ 0 \}$, then for any $x \neq 0$ $|f(x)-f(c)| = \left \lvert \frac{1}{x}- \frac{1}{c} \right \rvert = \frac{|x-c|}{|xc|}$Take $\delta_{1} = \frac{|c|}{2}$. Whenever $|x-c| < \delta_{1} \implies |c| -|x-c| > \frac{|c|}{2} \geq 0$, so using [[Reverse triangle inequality]], $|x|=|x-c+c| \geq ||c| - |x-c|| > \frac{|c|}{2}.$ Hence $|f(x)-f(c)| < \frac{2|x-c|}{|c^{2}|}$Given $\epsilon>0$, if also we have $|x-c|< \frac{|c^{2}|\epsilon}{2}$ then $|f(x) - f(c)| < \epsilon$Take $\delta=\min(\delta_{1}, \delta_{2})$, it follows that $x\in \mathbb{R} \backslash \{ 0 \}, |x-c|<\delta \implies |f(x) - f(c)|<\epsilon.$which shows that $f$ is continuous at $c$. **Proof (using [[Continuous Function (Sequential Continuity Definition)|sequence]])** Fix $c \neq 0$. Take $x_{n} \neq 0$ with $x_{n} \to c$. Then $f(x_{n}) = \frac{1}{x_{n}} \to \frac{1}{c} = f(c)$ as $n \to \infty$ using [[Algebra of Limits of Convergent Sequences|quotient rule for limits]]. So $f$ is continuous at $c$.