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

*Related*