Q:

If $f>0$ is continuous on a bounded closed interval, then $1/f$ is bounded there

Here's a problem I've been working on:
If $f$ is continuous on $[a, b]$ with $f(x)>0$ for all $a \le x\le b$, then $\frac 1f$ is bounded on $[a, b]$.
My thoughts so far: $1/f$ is bounded so long as $f$ doesn't approach $0$ since that is the only point at which $1/f$ will be undefined/unbounded. ( We know $f(x)$ can't ever be zero since $f(x) > 0$.)  Yet, f can never approach zero since x is bounded by a and b and thus there exists an $\epsilon>0$ such that for all $N$ in the natural numbers, there exists $n > N$ implying $|f(x-n)| \ge \epsilon$
Somehow, I do not feel this is nearly rigorous enough.  Yet, I am confused as to what else I can do.  Any suggestions are greatly appreciated!  

A:

Weierstrass' theorem ensures that $f$ reaches its maximum $M$ and its minimum $m$ in $[a,b]$. That is,
$$m\le f(x)\le M$$
thus
$$\frac1M\le\frac1{f(x)}\le\frac1m$$

