Proving a f is continuous at 1/3 Show that $f(x) = \frac{1}{5x}$ is continuous at $x = \frac{1}{3}$. I have to use an $\varepsilon - \delta$ proof? I am having trouble choosing my delta because I am confused as to how $|\frac{1}{x} - \frac13|$ can get to $|x - \frac{1}{3}|$?

A function is continuous at a given point $x_0$ if for every $ϵ>0$ you can find some $δ>0$ such that $|x−x_0|<δ⟹|f(x)−f(x_0)|<ϵ.$
What is useful with these problems is to "solve them backwards". Start with
$$\left|\frac{1}{5x} - \frac{1}{15}\right| < \epsilon$$
What happens when $0<x<3$?
$$\frac{1}{5x} - \frac{1}{15} < \epsilon$$
$$\frac{1}{x} - \frac{1}{3} < 5 \epsilon$$
$$\frac{1}{x} < \frac{1}{3} + 5 \epsilon = \frac{1+15 \epsilon}{3}$$
$$x > \frac{3}{1+15\epsilon}$$
What happens when $x>3?$
$$-\frac{1}{5x} + \frac{1}{15} < \epsilon$$
$$-\frac{1}{x} + \frac{1}{3} < 5\epsilon$$
$$  \frac{1}{3} - 5\epsilon <\frac{1}{x}$$
Now we're gonna want to guarantee that the left side of this inequality is positive, so choose $\epsilon <\frac{1}{15}.$
$$  \frac{1-15\epsilon}{3} <\frac{1}{x}$$
$$  \frac{3}{1-15\epsilon} >x$$.
In conclusion, we see that when $\epsilon <\frac{1}{15}$ and $x>0,$ then $x$ lies in the range
$$\frac{3}{1+15\epsilon} < x < \frac{3}{1-15\epsilon}.$$
So
$$\frac{3}{1+15\epsilon} -3 < x -3 < \frac{3}{1-15\epsilon} -3.$$
Now think about how you can choose $\delta$ so that $|x-3| < \delta$ implies the above condition. (Hint: just make sure $\delta$ is the minimum of the two). Do you need to worry about $x<0$? (Hint: you could just choose $\delta$ to guarantee this doesn't happen).
