I am trying to comprehend the epsilon-delta proofs and additive limit laws proofs and I have reached the following conclusions (A,B). If anyone could verify that my understanding on either is correct it would be very appreciated.
A. Let functions $f(x), t(x)$ with the same domain. Let $lim_{x\rightarrow a}f(x) = K$ and let $|f(x)-K|\ge|t(x)-T|$ for all $x$ in the domain. Show that $lim_{x\rightarrow a}t(x) = T$.
Proof:
Since $lim_{x\rightarrow a}f(x) = K$ we have that:
For any $\;\epsilon>0\;$ there exists $\;\delta>0\;$ s.t. if $\;0<|x-a|<\delta\;$ then $\;|f(x)-K|<\epsilon$
Since $|f(x)-K|\ge|t(x)-T|$ for all $x$ in the domain we have that:
For any $\;\epsilon>0\;$ there exists $\;\delta>0\;$ s.t. if $\;0<|x-a|<\delta\;$ then $\;|t(x)-T|<\epsilon$
Thus we reach: $lim_{x\rightarrow a}t(x) = T$
I understand this result as: anytime you want to show that $lim_{x\rightarrow a}t(x) = T$ it suffices to find any function $f(x)$ on the same domain, with $lim_{x\rightarrow a}f(x) = K$ and the property $|f(x)-K|\ge|t(x)-T|$ for all $x$ in the domain.
Is this correct? If so, can it be made to be more precise, i.e. does the property need to hold for all $x$ in the domain?
B. Let functions $f_1(x),\dots,f_n(x), t(x)$ with the same domain and $n$ finite.
Let $lim_{x\rightarrow a}f_i(x) = K_i$ for $i=1,\dots,n$ and let $\sum_{i=1}^n|f(x)-K_i|\ge|t(x)-T|$ for all $x$ in the domain.
Show that $lim_{x\rightarrow a}t(x) = T$.
Proof:
Since $lim_{x\rightarrow a}f_i(x) = K_i$ we have that:
For given $\;\epsilon/n\;$ there exists $\;\delta_1>0\;$ s.t. if $\;0<|x-a|<\delta_1\;$ then $\;|f_1(x)-K_1|<\epsilon/n$
$\vdots$
For given $\;\epsilon/n\;$ there exists $\;\delta_n>0\;$ s.t. if $\;0<|x-a|<\delta_n\;$ then $\;|f_n(x)-K_n|<\epsilon/n$
Let $\delta=min(\delta_1,\dots,\delta_n)$. We need to show that: if $\;0<|x-a|<\delta\;$ then $\;|t(x)-T|<\epsilon\;$
Since $0<|x-a|<\delta\le\delta_i$ for all $i=1,\dots,n$, all the following hold:
$\;|f_1(x)-K_1|<\epsilon/n\;$ and $\;|f_2(x)-K_2|<\epsilon/n\;$ and $\;\dots\;$ and $\;|f_n(x)-K_n|<\epsilon/n\;$
Thus if $0<|x-a|<\delta$ then $\sum_{i=1}^n|f_i(x)-K_i|<\epsilon$
Since $\sum_{i=1}^n|f_i(x)-K_i|\ge|t(x)-T|$ for all $x$ in the domain we have that:
if $\;0<|x-a|<\delta\;$ then $\;|t(x)-T|<\epsilon$
Thus we reach: $lim_{x\rightarrow a}t(x) = T$