if $g$ is continuous at $c$ and $g(c)\neq 0$, there exists an open interval containing $c$ on which $f(x)/g(x)$ is defined (Abbott p 113 q4.3.5)

Question

Theorem 4.3.4.(iv) says that $f(x)/g(x)$ is continuous at $c$ if both $f$ and $g$ are, provided that the quotient is defined. Show that if $g$ is continuous at $c$ and $g(c)\neq 0$, then there exists an open interval containing $c$ on which $f(x)/g(x)$ is always defined.

Solutions to Homework 6 Let $A$ be the domain of $f$. Define $e =\displaystyle \frac{1}{2}|g(c)|$.

1. How can we presage this choice of e? I don't understand the modus operandi of this proof.

2. Intuition please?

Then $e >0$. $g$ is posited as continuous at $c$, so choose $\delta>0$ such that for all $x\in A$ with $|x-c|<\delta$, we have $|g(x)-g(c)|< e$. For $x\in(c-\delta,\ c+\delta)\cap A$, use the triangle inequality at the first step:

3. Don't we write $|x -c | <\delta $ for delta-epsilon proofs? Why $x\in(c-\delta,\ c+\delta)\cap A$?

4. The reverse triangle inequality is $|g(x) - g(c)| \ge \color{red}{|} \, |g(x)| - |g(c)| \, \color{red}{|}$. How does this invoke:

$ \begin{align} |g(x)| & \geq |g(c)|-|g(x)-g(c)| \\ & >|g(c)|-\xi \\ & =|g(c)|-\frac{1}{2}|g(c)| \\ & >0. \end{align} $

Answer 1

1. There is no reason to pick $e=\frac12|g(c)|$. Apparently this choice is just based on the intuition that "$\frac\epsilon2$ always works, and if it doesn't, try $\frac\epsilon3$". Picking $e=|g(c)|$ suffices (try it with the rest of the proof!)

2. The intuition behind picking $e=|g(c)|$? After all, $|g(c)|$ is about the only number we know from the probem statement that is positve ...

3. $x\in(c-\delta,c+\delta)\cap A$ is equivalent to $|x-c|<\delta$ and $x\in A$.

4. Note that $|a|\ge a$. Also, you can use the usual triangle inequality $|g(c)|=|g(x)+(g(c)-g(x))|\ge |g(x)|+|g(c)-g(x)|=|g(x)|+|g(x)-g(c)|$.

Answer 2

We know that $g(c)\ne0$, and we are trying to find an $\epsilon>0$ such that the interval $(g(c)-\epsilon,g(c)+\epsilon)$ does not contain 0.

Since $\vert g(c)\vert$ represents the distance from $g(c)$ to 0, we can take $\epsilon$ to be any number satisfying $0<\epsilon\le\big\vert g(c)\big\vert$ to avoid having 0 in this interval, and they just chose $\epsilon=\frac{|g(c)|}{2}$ for convenience.

If $\vert g(x)-g(c)\vert<\epsilon$, we want to show that $g(x)\ne0$; and this is true since $g(x)=0\implies \big\vert-g(c)\big\vert<\epsilon\implies \big\vert g(c)\big\vert<\epsilon$, which contradicts our choice of $\epsilon$.

We can also show this using the Reverse Triangle Inequality as they do in the proof, though, since $|g(x)|=\big\vert g(c)-(g(c)-g(x))\big\vert\ge\big\vert|g(c)|-|g(c)-g(x)|\big\vert\ge|g(c)|-|g(c)-g(x)|=|g(c)|-|g(x)-g(c)|>|g(c)|-\epsilon\ge0,$

so $\vert g(x)\vert>0$ and therefore $g(x)\ne0$.