Proof of the Sum Law for Limits

The Sum Law for Limits

If $\lim_{x \rightarrow c} f(x)$ and $\lim_{x \rightarrow c} g(x)$ both exist, then $$\lim_{x \rightarrow c} (f(x) + g(x)) = \lim_{x \rightarrow c} f(x) + \lim_{x \rightarrow c} g(x)$$ (This is often loosely stated as "the limit of a sum is the sum of limits".)



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.