In this post, I show that $ \lim_{x \rightarrow a} [f(x) + g(x)]$ is equal to $\lim_{x \rightarrow a}f(x) + \lim_{x \rightarrow a} g(x)$ given that $ \lim_{x \rightarrow a}f(x) = A$ and $ \lim_{x \rightarrow a}g(x) = B$. To do this, I approximately follow the steps in Reference [1]. Objective: Using the definition of a limit, the objective is to show that: For each positive number $\epsilon_2$ there is another, $\delta$, such that whenever $ 0 < |x-a| < \delta $ we have $ |f(x)+g(x) – (A+B)| < \epsilon_2$. Proof: From the definition of a limit, it is known that For each positive number $\epsilon_0$ there is another, $\delta$, such that whenever $ 0 < |x-a| < \delta $ we have $ |f(x) – A| < \epsilon_0$. For each positive number $\epsilon_1$ there is another, $\delta$, such that whenever $ 0 < |x-a| < \delta $ we have $ |g(x) – A| < \epsilon_1$ Consider the expression, $|f(x)+g(x) – (A+B)|$. This can be rearranged: $ |f(x)+g(x) – (A+B)| = |f(x)-A+g(x)-B| $. Next, use the triangle inequality to write $ |f(x)-A+g(x)-B| \le |f(x)-A|+|g(x)-B| $. Using what is known about the individual limits, there is an upper bound: $ |f(x)-A|+|g(x)-B| < \epsilon_0 + \epsilon_1$. Define $\epsilon_2 \equiv \epsilon_0 + \epsilon_1$. Since $ \epsilon_0 > 0 $ and $ \epsilon_1 > 0 $, it follows that $\epsilon_2 > 0$ as needed. Therefore, $|f(x)+g(x) – (A+B)| \le |f(x)-A|+|g(x)-B| < \epsilon_2$. Since there is no equation relating $\epsilon_2$ and $\delta$, $|f(x)+g(x) – (A+B)| < \epsilon_2$ whenever $ 0 < |x-a| < \delta $ with $\epsilon_2 > 0$. $\square$ References: [1] https://tutorial.math.lamar.edu/classes/calci/limitproofs.aspx