continuity with epsilon delta for $f(x) = 1/g(x)$

Question

Let $g\colon\mathbb{R}\longrightarrow\mathbb R$ be a continuous function, where $g(x)>0$ for all $x$. Determine continuity with delta epsilon for function $f$ where $f(x) = 1/g(x)$ and $x\ge1$.

How to proof that this function is continuous?

Answer 1

You start like you always do when proving something is continuous from the definition: pick an arbitrary point $p$, and an $\epsilon>0$, and work your way from there.

You will need to use the fact that $g$ is continuous, of course. Also, related to that, it usually makes things easier if you put some bound on what $\delta$ you're allowing yourself to use ahead of time. When proving that $x^2$ is continuous, for instance, one usually decides that $\delta<1$ in order to get control over certain terms later on. It's not quite as simple in this case, but it works on the same principle.

This is all I'm willing to give you for starters. Give it a try, and post the results as an edit to your question. I might not have time to help any more today, since I have other plans, but I'll check in tomorrow if I can. Or someone else might help you before then once you get started.

Answer 2

Let $a\ge1$ (we don't need this assumption actually). Since $g$ is continuous, $\exists d>0$ such that $\forall x\in(a-d,a+d)$, $|g(x)-g(a)|<g(a)/2$, i.e.

$$\frac{g(a)}{2}<g(x)<\frac{3g(a)}{2}$$

Let $\epsilon>0$. $\exists \delta_1>0$ such that $\forall x\in(a-\delta_1,a+\delta_1)$,

$$|g(x)-g(a)|<\frac{\epsilon g(a)^2}{2}$$

Let $\delta=\min\{d,\delta_1\}$. Then $\forall x\in(a-\delta,a+\delta)$, $\displaystyle |g(x)|=g(x)>\frac{g(a)}{2}$ and

\begin{align} |f(x)-f(a)|&=\left|\frac{g(a)-g(x)}{g(x)g(a)}\right|\\ &<\left| \frac{\displaystyle\frac{\epsilon g(a)^2}{2}}{\displaystyle\left(\frac{g(a)}{2}\right)g(a)}\right|\\ &=\epsilon \end{align}

So $f$ is continuous at $x=a$.