Show that $f(x)=1/ x$ is continuous at any $c\neq 0$
Solution 1:
Suppose $c>0$. If $x > \frac{c}{2}$, then $|\frac{1}{x}-\frac{1}{c}|= \frac{|x-c|}{xc} < \frac{2}{c^2}|x-c|$
So, if $\epsilon>0$, choose $\delta < \min(\frac{c^2\epsilon }{2},\frac{c}{2})$, then if $|x-c|< \delta$, we have (i) $x > \frac{c}{2}$, and from above we have (ii) $|\frac{1}{x}-\frac{1}{c}| < \frac{2}{c^2}\delta \le \epsilon $.
A similar argument applies to $c<0$. Or you could use the fact that $f(x) = -f(-x)$, and use the fact that multiplication by a constant ($-1$, in this case) is continuous, and composition of continuous functions is continuous.