calculus - Any proof that verify why the limit of the difference is the difference of the limits?

I did a research on internet and books about why the difference of the limits is the difference of the limits, but i didn't get any result of this proof. I would appreciate if somebody can help me. Thanks. :)

Answer

Consider $f(x)$ and $g(x)$ are two functions on real domain whose limits exist for $x=a$.

Given: $\lim_{x \to a} f(x)$ = $K$ and $\lim_{x \to a} g(x)$ = $L$

Let $\varepsilon$ > 0 then there exist $\alpha$ > $0$ and $\beta$ > $0$ such that ,

$|f(x)-K |$< $\varepsilon$/2 ,whenever $0$<|$x$-$a$|<$\alpha$ ,and
$|g(x)-L |$< $\varepsilon$/2 , whenever $0$<|$x$-$a$|<$\beta$

Choose $\gamma$= $min${$\alpha$,$\beta$}

Now we need to show that
$|f(x)+g(x)-(K+L)|$

Assume that we have $0$<$|x-a|$<$\gamma$. Then we have,

$|f(x)+g(x)-(K+L)|$=$|(f(x)-K)+(g(x)-L)|$ < $|(f(x)-K)|$ + $|(g(x)-L)|$ < $\varepsilon$

In our third step we used the fact that, by our choice of $\gamma$, we also have $0$<|$x$-$a$|

So we can use initial statements in our proof.

Now replace $g(x)$ by $(-1)g(x)$ and you will get your proof.