Here's the question:
Let $f$ be a function. If $f(x) = k$, where $k$ is a constant, in some deleted neighbourhood of $x_0$, then:
$$\lim_{x \to x_0} f(x) = k$$
Proof Attempt:
Let $\delta_1 > 0$ exist so that:
$$0 < |x-x_0| < \delta_1 \implies f(x) = k$$
Now, let $\epsilon > 0$ be given. Then, pick $\delta = \delta_1$. So, it is the case that:
$$0 < |x-x_0| < \delta_1 \implies |f(x)-k| = 0 < \epsilon$$
The statement above holds trivially because $\epsilon > 0$ is true and, therefore, the implication is always true. Hence:
$$\lim_{x \to x_0} f(x) = k$$
Does the proof above work? If it doesn't, how can I fix it?