One of my Calculus profs recorded a video (screenshot below) to prove the limit law for the sums of functions:
$$\lim_{x\to a} f(x) = L \ \wedge \ \lim_{x\to a} g(x) = M \Longrightarrow \lim_{x\to a} (f(x) + g(x)) = L + M$$
But I think he made a small mistake.
I think he should've taken $\delta = \max \{\delta_1, \delta_2\}$ instead of $\min \{ \delta_1, \delta_2\}$.
Is that right, or am I misunderstanding something?
(The small text in green, blue, and orange is my addition.)
On
It should be $\min$:
We want $|f(x)-L|<\frac{\epsilon}{2}$, and this only happens when $|x-a|<\delta_1$. Similarly, we want $|g(x)-L|<\frac{\epsilon}{2}$, which only happens when $|x-a|<\delta_2$.
Thus we want $|x-a|$ to be smaller than both $\delta_1$ and $\delta_2$, that is, $x<\min(\delta_1,\delta_2)$. Thus we set $\delta=\min(\delta_1,\delta_2)$.
If $|x-a|$ were less than only the maximum of $\delta_1,\delta_2$ then it is possible that $\delta_1<x<\delta_2$, so $|f(x)-L|>\frac{\epsilon}{2}$ which is obviously bad for the proof.
I hope this clears things up :)
No, $\min$ is correct. Let's take your example numbers just to have something concrete to work with.
If $0<|x-a|<2$, then clearly $0<|x-a|<3$ also holds. So the hypothesis for (2) holds, and we are good to go.
On the other hand, if we only require $0<|x-a|<3$, then we do not neccessarily have $0<|x-a|<2$, so the hypothesis of (1) doesn't hold, and thus we can't conclude that $|f(x)-L|<\varepsilon/2$.
Basically, given our $\varepsilon$, we have found several $\delta_i$'s, each one restricting what $x$ can be for that specific part of the argument to hold. We pick the most restrictive of all the $\delta_i$ in order to ensure that all of the hypotheses hold simultaneously, which again mean that all the conclusions hold, and we can go on to complete our proof.