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Notes:

Considering two limit were given

$$\mathop {\lim }\limits_{x \to a} f\left( x \right) = L \hspace{0.1in} and \hspace{0.1in} \mathop {\lim }\limits_{x \to a} g\left( x \right) = M$$

means there is $\delta_1>0$ so that$ |f(x)-L|< \varepsilon$ whenever $<0|x-a|>\delta_1$

also there is $\delta_2>0 $ so that $|g(x)-M|< \varepsilon$ whenever $<0|x-a|>\delta_2$

choose $\delta=\min(\delta_1,\delta_2)$ then

$|f(x)-L|< \varepsilon$ and $|g(x)-M|< \varepsilon$ whenever $0<|x-a|<\delta$

Question:

Is it okay Substitute "and" with "+" so that

$|f(x)-L| + |g(x)-M|< \varepsilon + \varepsilon$ whenever $0<|x-a|<\delta$

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  • $\begingroup$ If you choose $\delta$ so $|f(x)-L|$ and $|g(x)-M|<\dfrac\varepsilon 2$ then $|f(x)+g(x)-(L+M)|<\varepsilon$ $\endgroup$ – J. W. Tanner May 25 at 21:50
  • $\begingroup$ The symbol $\land$ can be used to mean “and” $\endgroup$ – J. W. Tanner May 25 at 21:53
  • $\begingroup$ @J.W.Tanner Putting $\dfrac\varepsilon 2$ is helping when i proof the sum theorem but i just want to put two epsilons in one side by adding $|f(x)-L|+|g(x)-M|<\varepsilon + \varepsilon$ instead of $|f(x)-L|< \varepsilon $and$ |g(x)-M|<\varepsilon$ $\endgroup$ – Ammar Bamhdi May 25 at 22:08
  • $\begingroup$ If $|f(x)-L|<\varepsilon$ and $|g(x)-M|<\varepsilon$ then $|f(x)-L|+|g(x)-M|<\varepsilon+\varepsilon=2\varepsilon$, if that’s what you’re asking now $\endgroup$ – J. W. Tanner May 25 at 22:57
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    $\begingroup$ Yes, if $a<b$ and $c<d$ then $a+c<b+c<b+d$ $\endgroup$ – J. W. Tanner May 25 at 23:35
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Answer to the edited question:

If $|f(x)-L|<\varepsilon$ and $|g(x)-M|<\varepsilon,$ then it is valid to conclude that $|f(x)-L|+|g(x)-M|<\varepsilon+\varepsilon=2\varepsilon.$ This is an application of $a<b$ and $c<d \implies a+c<b+d$, which follows from $a+c<b+c$ and $b+c<b+d$.

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No, absolutely not. I get where you're coming from with the "$+$" as a substitute for "and" but it makes it look weirder, if not altogether wrong. For example if I had the two statements

$$1 < 2 \;\;\;\;\; \text{and} \;\;\;\;\; -3 < 1$$

but did this replacement here I would have

$$1 < 2 + -3 < 1$$

which looks like an intermediate step in saying $1<-1<1$ which is just plain nonsense.

An argument could be made for using an ampersand instead, "&", but I think something important to bear in mind is that in proofs you're usually expected to use words. Once you start doing proof-based things, you're also having to communicate ideas as well as the underlying mathematics, and for that you might want to write in prose as opposed to going down to shorthand. Not just to avoid ambiguities like the above but also just in the interest of communication.

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    $\begingroup$ +1 for the general recommendation to "use words". $\endgroup$ – Ethan Bolker May 25 at 21:54
  • $\begingroup$ Hi Eevee Trainer thanks for helping, in fact i was wondering if i can do this to combine $|f(x)-L| + |g(x)-M|< \varepsilon+ \varepsilon$ instead of $|f(x)-L|<\varepsilon $ and $|g(x)-M|< \varepsilon$ $\endgroup$ – Ammar Bamhdi May 25 at 21:59
  • $\begingroup$ I mean, I guess you could. That said whether it could be a good idea to do so depends on the context and what you're trying to do. $\endgroup$ – Eevee Trainer May 25 at 22:02
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    $\begingroup$ @AmmarBamhdi Be aware that $$|f(x) - L| + |g(x) - M| < \varepsilon + \varepsilon$$ and $$|f(x) - L| < \varepsilon \text{ and } |g(x) - M| < \varepsilon$$ are not equivalent! The latter implies the former, but not the other way around. It's a valid logical step to conclude the former given the latter, but they do not mean the same thing! $\endgroup$ – Theo Bendit May 25 at 22:16
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    $\begingroup$ @TheoBendit- I can understand that we can go from $|f(x)−L|<ε and |g(x)−M|<ε$ to $|f(x)−L|+|g(x)−M|<ε+ε$ but not the revers $\endgroup$ – Ammar Bamhdi May 25 at 22:24
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No, do not do this. Write "and". Why would you need to save 2 letters anyway?

In fact, $$ |f(x)-L|< \varepsilon + |g(x)-M|< \varepsilon $$ already has a meaning. It represents two inequalities: $$ |f(x)-L|< \varepsilon + |g(x)-M|\qquad \text{and} \qquad \varepsilon + |g(x)-M| < \varepsilon $$

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  • $\begingroup$ Hi GEdgar thanks for helping, I think i made a mistake, i want to reach $|f(x)-L|+|g(x)-M|<\varepsilon + \varepsilon$ and not $|f(x)-L|< \varepsilon + |g(x)-M|< \varepsilon$ by replace "and" with "+" in $|f(x)−L|<ε$ and $|g(x)−M|<ε$ $\endgroup$ – Ammar Bamhdi May 25 at 22:12

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