Show $f(x) = \frac{1}{x}$ is continuous at $x = \frac{1}{2}$.

Question

The definition of continuity we're trying to satisfy is $\forall \epsilon > 0, \exists \delta > 0$ such that, $\forall |x - \frac{1}{2}| < \delta$, we have $|f(x) - f(\frac{1}{2})| < \epsilon$.

So in this case, we let $\epsilon > 0$ be given and choose a $\delta$ (I'm not sure what the value should be. $\forall|x - \frac{1}{2}| < \delta$, we have:

$|f(x) - f(\frac{1}{2})| = |\frac{1}{x} - \frac{1}{\frac{1}{2}}| = |\frac{1}{x} - 2|$.

I am lost on where to go from here. I'm assuming $\delta$ will be equal to $c\epsilon$ for some constant c, and if I can reduce the above expression to something $< \frac{|x - \frac{1}{2}|}{c}$, I'll be done, but I'm not sure at all what steps to take.

Answer 1

\begin{align} \left|\frac1x - \frac1{\frac12} \right| &=\left|\frac{x-\frac12}{\frac{x}2} \right| \end{align}

Let's try to have some control over the magnitude of $\frac{x}2$.

If we pick $\delta <\frac14$, then $|x-\frac12| < \frac14$ would imply that $$\frac14<\left| x\right| < \frac34.$$

Are you able to solve the problem now?

Answer 2

Use the algebraic continuity theorem. Since we want to prove that $f(x)$ is continuous at $x=\frac12$, we can equivalently prove that $1/f(x)$ is continuous at $x=\frac12$ and that $f(x)\neq 0$ at $\frac12$.