Is there an error in this epsilon-delta proof of the limit law for the sums of functions

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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?

Screenshot of the proof from the video. The small text in green, blue, and orange is my addition.

(The small text in green, blue, and orange is my addition.)

Best Answer

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 :)

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