Prove if $f(x)$ and $g(x)$ is continuous, then $f(x) + g(x)$ is also continuous using the $\epsilon - \delta$ definition of limits

Since both $f(x)$ and $g(x)$ is continuous, then $$(\forall \epsilon_1 >0)(\exists \delta_1 >0) [\vert x-a\vert< \delta_1 \to \vert f(x)-f(a) \vert <\epsilon_1]$$$$(\forall \epsilon_2 >0)(\exists \delta_2 >0) [\vert x-a\vert< \delta_2 \to \vert g(x)-g(a) \vert <\epsilon_2]$$

I defined $h(x)=f(x)+g(x)$ and for a given $\epsilon>0$, I set both $\epsilon_1,\epsilon_2=\frac12\epsilon$.

Could I ask why would we set $\epsilon_1,\epsilon_2=\frac12\epsilon$? I mean could we can also set it as $\epsilon_1=\frac14\epsilon$ and $\epsilon_2=\frac34\epsilon$?

Additionally, after setting this, how did we derive $\delta=min\{\delta_1, \delta_2\}$, which is required for the proof?

• You could set $\epsilon_1=\epsilon/4$ and $\epsilon_2=3\epsilon/4$ without any problems. You just have to make a choice, and $\epsilon/2$ for both of them looks nice. Jun 21 '16 at 17:00
• Why use those choices of $\epsilon$ and $\delta$?Because it's the most straightforward choice to get a sum to $\epsilon$. You could certainly use other values,just like you can take other routes to get to a certain destination. Jun 21 '16 at 17:06

Yes, you could also choose $\epsilon_1 = \frac{1}{4}\epsilon$ and $\epsilon_2 = \frac{3}{4}\epsilon$, because you just need $$\lvert f(x) + g(x) - (f(a) + g(a))\rvert \leq \lvert f(x) - f(a)\rvert + \lvert g(x) - g(a)\lvert$$ to be less than $\epsilon$. In fact you just need $\epsilon_1 + \epsilon_2 \leq \epsilon$.

By picking $\delta$ the smallest of the two deltas, you guarantee that both $\lvert f(x) - f(a)\rvert < \epsilon_1$ and $\lvert g(x) - g(a)\lvert < \epsilon_2$.

Ok, to see that $h(x)=f(x)+g(x)$ is continuous you have to see that given $\epsilon_{0} >0$ there exist $\delta>0$ such that $|x-a|<\delta$ implies $|h(x)-h(a)|<\epsilon$. Now we know that

$$|h(x)-h(a)|=|f(x)+g(x)-f(a)+g(a)|\leq|f(x)-f(a)|+|g(x)-g(a)|$$

because of the triangle inequality. So we already have our given $\epsilon_{0}>0$ we want to find the $\delta_{0}$. So now we consider $\epsilon_0/2$ (now in a minute you will se why). Since both $f$ and $g$ are continuous, there exit a $\delta_1$ and $\delta_2$ such that

$$|x-a|<\delta_1 \implies |f(x)-f(a)|<\epsilon_0/2$$

and

$$|x-a|<\delta_2 \implies |g(x)-g(a)|<\epsilon_0/2$$

So if you take $\delta_0=min\{\delta_1,\delta_2\}$ any time $|x-a|<\delta_0$ both of the above will be true. But if both of the above are true, we have that

$$|h(x)-h(a)|\leq|f(x)-f(a)|+|g(x)-g(a)|<\epsilon_0/2+\epsilon_0/2 = \epsilon_0$$

And the proposition is proved.