The proposition starts with suppose that the limit of $f(x)=a$ as $x\to{p}$ and the limit of $g(x)=b$ as $x\to{p}$.
I know we have to start of with the definition of what the prop starts of with. But then I don't know how to proceed from there.
The proposition starts with suppose that the limit of $f(x)=a$ as $x\to{p}$ and the limit of $g(x)=b$ as $x\to{p}$.
I know we have to start of with the definition of what the prop starts of with. But then I don't know how to proceed from there.
Given. We have that for all $\epsilon > 0$, there exists a $\delta > 0$ such that $\delta > |x-p| > 0$ implies:
$$ |f(x) - a| < \epsilon/2$$ $$ |g(x) - b| < \epsilon/2$$
Rest of the proof. Adding these two inequalities,
$$|f(x) - a| + |g(x) - b| < \epsilon.$$
By the Triangle Inequality, the LHS is greater than $|(f(x)-a) + (g(x) - b)| = |(f(x)+g(x)) - (a+b)|$, and so
$$|(f(x)+g(x)) - (a+b)| < \epsilon.$$
So, for all $\epsilon > 0$, there exists an $\delta > 0$ such that $\delta > |x-p| > 0$ implies
$$|(f(x)+g(x)) - (a+b)| < \epsilon.$$
Thus the limit as $x \rightarrow p$ of $(f+g)(x) = a+b$.