Let $f$ be a function. Suppose that the value of $f$ at $x_1 \neq x_0$ is changed and also suppose that $\lim_{x \to x_0} f(x) = L$ before the change. Then, $\lim_{x \to x_0} f(x) = L$ after the change.
Proof Attempt:
Let $f(x_1) = a$ originally and let it now be set to $f(x_2) = b$. Suppose that doing this changed the limit at $x_0$ to $L'$.
Then, we know that there exists $\delta_1,\delta_2>0$ such that:
$$0<|x-x_0| < \delta_1 \implies |a-L| < \epsilon$$
$$0<|x-x_0| < \delta_2 \implies |b-L'| < \epsilon$$
Let $\delta = \min\{\delta_1,\delta_2\}$. Then, it is the case that:
$$|(a-b)+(L'-L)| \leq |a-L| + |b-L'| < 2\epsilon$$
Then, set $\epsilon = \frac{1}{2} \cdot |(a-b)+(L'-L)|$. This brings about a contradiction. So, $L = L'$ and that proves the desired result.
Does the proof above work? If it doesn't, how can I fix it?