# The Limit Sum Rule $ \lim_{x \to a} {[f(x) + g(x)]} = \lim_{x \to a} {f(x)} + \lim_{x \to a} {g(x)} $ ^wjie2a ## Proof Our goal is to evaluate $\lim_{x \to a} [f(x) + g(x)]$ in terms of $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$. Put in other words: we’re looking for an $M$ such that for any given $\epsilon$ we can come up with a range of values that satisfy the following inequality: $f(x) + g(x) - M < \epsilon$. $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = K$ imply that we can define, of our own volition, two values $\epsilon_L$ and $\epsilon_K$ such that $|f(x) - L| < \epsilon_L$ and $|g(x) - K| < \epsilon_K$ for a given range of $x$ values. Note that that the two limits will yield two different $\delta$ values for each, but we can always take the smaller one, and we’ll get a range where both inequalities are satisfied. Now we’ll add the two inequalities together: $ \begin{gather} |f(x) - L| + |g(x) - K| < \epsilon_L + \epsilon_K & \text{per triangle inequality} \\ |f(x) - L + g(x) - K| \le|f(x) - L| + |g(x) - K| < \epsilon_L + \epsilon_K & \text{we can remove the middle part} \\ |f(x) - L + g(x) - K| < \epsilon_L + \epsilon_K \\ |f(x) + g(x) - (L + K)| < \epsilon_L + \epsilon_K \end{gather} $ Because we are free to define $\epsilon_L$ and $\epsilon_K$ we can define them such that their sum is $\epsilon$. $ |f(x) + g(x) - (L + K)| < \epsilon $ This implies that $M = L + K$ or $\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$. See [[Absolute Value#The Triangle Inequality]] for more about the triangle inequality.