$$\lim _{x\rightarrow a} f(x)+g(x) = f(a)+g(a)$$
Let $\epsilon>0$ be given
Since $f$ and $g$ are continuous, $|f(x)-f(a)|< \epsilon$ when $0<|x-a|< \delta_f $ and $|g(x)-g(a)|< \epsilon$ $\ $ when $0<|x-a|< \delta_g$
Let $\delta_h$ be defined as $\min(\delta_g ,\delta_f)$ and $h(x)$ be defined as $f(x)+g(x)$
$$\ |f(x)-f(a)|+|g(x)-g(a)|< 2\epsilon\tag{1}$$
$$\ |f(x)-f(a) + g(x)-g(a)|< 2\epsilon$$
$$\ |h(x)-h(a)|< 2\epsilon$$
We can replace $\delta_f$ and $\delta_g$ by
$𝛿_h$ to get
$|f(x)-f(a)| < \epsilon$ when $0<|x-a|<\delta_h$ and $\ |g(x)-g(a)|< \epsilon$ when $0<|x-a|<\delta_h$
Since (1) is true when $0<|x-a|<\delta_h$, we can find a $\delta_h$ for every value of $2ε$
for every $2\epsilon>0$ there is a $\delta>0$ such that $|x−a|<\delta\implies ∣h(x)−h(a)∣<2\epsilon$ and for every $2\epsilon$ there exists an $\epsilon$ so for every $\epsilon$ there exists a $\delta$.
Is the proof logically correct and is it worded correctly?
Best Answer
It is quite good, but no totally correct. You should be aiming at the inequality$$|h(x)-h(a)|<\varepsilon.\tag1$$So, your numbers $\delta_f$ and $\delta_g$ should be such that$$|x-a|<\delta_f\implies|f(x)-f(a)|<\frac\varepsilon2$$and that$$|x-a|<\delta_g\implies|g(x)-g(a)|<\frac\varepsilon2.$$Then, yes, if you take $\delta=\min\{\delta_f,\delta_g\}$ and if $|x-a|<\delta$, then $(1)$ holds.