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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 $\lim_{x \rightarrow c} f(x)$ 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.