Derivation of $|f(x)-f(x_{0})|<\frac{\epsilon}{2(|g(x_{0})|+\epsilon+1)}\;\;\forall\; x\;:\;|x-x_{0}|<\delta_{1}$

Question

I am reviewing the product theorem of continuity. Here is the theorem

Let $f$ and $g$ be continuous at $x_{0}$ belonging to the intersection of the domians of $f$ and $g$. Then, I'm to prove that $fg$ is continuous at $x_{0}$, where $$(fg)(x)=f(x)g(x)\;\; \forall \;x\,\in D(f)\cap D(g).$$

Here's a part of my proof!

Since $f$ is continuous at $x_{0}$, then for any $\epsilon>0$ there exists a $\delta_1>0$ such that

$$|f(x)-f(x_0)|<\epsilon\;\; \text{whenever} \;\;|x-x_0|<\delta_1.\qquad\qquad (1)$$

Also, since $g$ is continuous at $x_{0}$, then for any $\epsilon>0$ there exists a $\delta_2>0$ such that

$$|g(x)-g(x_0)|<\epsilon\;\; \text{whenever} \;\;|x-x_0|<\delta_2.\qquad\qquad (2)$$

Now,

$$|(fg)(x)-(fg)(x_0)|\leq |g(x)[f(x)-f(x_0)]| +|f(x_0)[g(x)-g(x_0)]|$$ $$\leq |g(x)|\cdot|f(x)-f(x_0)| +|f(x_0)|\cdot|g(x)-g(x_0)| .$$

My question is: How did the author modify $(1)$ and $(2)$ to give
$$|f(x)-f(x_0)|<\dfrac{\epsilon}{2(|g(x_0)|+\epsilon+1)}\;\; \forall\;x:\;|x-x_0|<\delta_1,$$ and $$|g(x)-g(x_0)|<\dfrac{\epsilon}{2(|f(x_0)|+\epsilon+1)}\;\; \forall\;x:\;|x-x_0|<\delta_2?$$.

Any help on this will be highly appreciated!

Answer 1

There is no modification.

For any given $\epsilon$, just treat $\dfrac{\epsilon}{2(|g(x_0)|+\epsilon+1)}$ as a new $\epsilon$ and "update" your $\delta$ choices accordingly.

Answer 2

@ Argyll: Are you sure of that because from $(1)$ and $(2)$

$$|f(x)-f(x_0)||g(x)-g(x_0)|<\epsilon^2.$$ This implies that $$|f(x)-f(x_0)||g(x)-g(x_0)|\leq |g(x)|\cdot|f(x)-f(x_0)| +|g(x_0)|\cdot|f(x)-f(x_0)|<\epsilon^2 $$ But $$|g(x)|=|g(x)-g(x_0)+g(x_0)|\leq |g(x)-g(x_0)|+|g(x_0)| $$ $$< \epsilon+|g(x_0)|< \epsilon+1 ,\;\;\text{if}\;\;|g(x_0)| <1.$$

Thus, $$(\epsilon+1)\cdot|f(x)-f(x_0)| +|g(x_0)|\cdot|f(x)-f(x_0)|<\epsilon^2 $$ $$\implies |f(x)-f(x_0)|<\dfrac{\epsilon^2}{|g(x_0)|+\epsilon+1} \;\;\text{whenever}\;\;|x-x_0|<\delta_1.$$

My result does not tally with the authors' result. Do I also say I am correct?