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We won't try to prove each of the limit laws using the epsilon-delta definition for a limit in this course. Despite the fact that these proofs are technically needed before using the limit laws, they are not traditionally covered in a first-year calculus course. Instead, they typically show up in a course in Mathematical Analysis that comes later. (I suppose the effect is similar to the filming of a movie prequel.)
Still, it is important that students understand how the epsilon-delta definition is the foundation on which the limit laws are built.
As such, we will look at just one of the limit laws (i.e., the limit of sum is a sum of limits), and how one can be assured it holds because of the epsilon-delta definition of a limit.
Here's the formal statement of this limit law and its proof:
If lim and \lim_{x \rightarrow c} g(x) both exist, then \displaystyle{\lim_{x \rightarrow c} (f(x) + g(x)) = \lim_{x \rightarrow c} f(x) + \lim_{x \rightarrow c} g(x)}
Proof:
Since we are working under an assumption that all of the limits above exist, let us suppose \lim_{x \rightarrow c} f(x) = L_1 and \lim_{x \rightarrow c} g(x) = L_2. So we now hope to show that \lim_{x \rightarrow c} (f(x) + g(x)) = L_1 + L_2 Appealing to the epsilon-delta definition for a limit, we must show that for any \epsilon > 0, we can find a \delta > 0 such that \textrm{If $0 < |x-c| < \delta$, then $|(f(x)+g(x))-(L_1+L_2)| < \epsilon$} Let's look at what we know...
We know \lim_{x \rightarrow c} f(x) = L_1. So -- again appealing to the epsilon-delta definition, only this time in reverse -- we also know that for any \epsilon_1 \gt 0, we can find \delta_1 \gt 0 such that if 0 < |x-c| < \delta_1, then |f(x)-L_1| \lt \epsilon_1.
We also know \lim_{x \rightarrow c} g(x) = L_2. So -- again appealing to the epsilon-delta definition, only this time in reverse -- we also know that for any \epsilon_2 \gt 0, we can find \delta_2 \gt 0 such that if 0 \lt |x-c| \lt \delta_2, then |g(x)-L_2| \lt \epsilon_2.
Consider what happens when we take both \epsilon_1 and \epsilon_2 to be \frac{\epsilon}{2}.
There must consequently be a \delta_1>0 and a \delta_2>0 such that: \textrm{If $0 < |x-c| \lt \delta_1$, then $|f(x)-L_1| \lt \frac{\epsilon}{2}$} \textrm{and} \textrm{If $0 < |x-c| \lt \delta_2$, then $|g(x)-L_2| \lt \frac{\epsilon}{2}$} Suppose \delta was the smaller of the two values, \delta_1 and \delta_2.
First notice it must be the case that \delta>0, as d_1,d_2>0.
Further, it should also then be true that \textrm{If $0 \lt |x-c| \lt \delta$, then $|f(x)-L_1| \lt \frac{\epsilon}{2} \quad \textrm{and} \quad |g(x)-L_2| \lt \frac{\epsilon}{2}$}
Adding the inequalities together, we find \textrm{If $0 \lt |x-c| \lt \delta$, then $|f(x)-L_1| + |g(x)-L_2| \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$} We can use the triangle inequality for absolute values (i.e. |A+B| \le |A| + |B|) to discover \textrm{If $0 \lt |x-c| \lt \delta$, then $|f(x)-L_1+g(x)-L_2| \le |f(x)-L_1| + |g(x)-L_2| \lt \epsilon$}
Finally, rearranging some terms inside one of the absolute values, and removing the now superfluous sum of absolute values to it's right, we find \textrm{If $0 \lt |x-c| \lt \delta$, then $|(f(x)+g(x))-(L_1+L_2)| \lt \epsilon$} which is what we hoped to show. Thus, \lim_{x \rightarrow c} (f(x) + g(x)) = L_1 + L_2 and consequently, \lim_{x \rightarrow c} (f(x) + g(x)) = \lim_{x \rightarrow c} f(x) + \lim_{x \rightarrow c} g(x) Q.E.D.