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 $0<|x-a|<\delta$
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 $b+c<b+d$.