So, here's what I'm trying to prove:
Let $f$ and $g$ be functions. Suppose that $f(x) \to c$ as $x \to x_0$ and $g(x) \to \infty$ as $x \to x_0$. Suppose that $c$ is finite. Then, $f(x)+g(x) \to \infty$ as $x \to x_0$.
Proof Attempt:
Let $M>0$. We have to prove that there is a $\delta>0$ such that:
$$0 < |x-x_0| < \delta \implies |f(x)+g(x)| > M$$
Now, we know that:
$$|f(x)+g(x)| \geq |g(x)|-|f(x)|$$
and we want that right-hand side to be greater than $M$. In other words, $|g(x)| > |f(x)| + M$. Let $\epsilon > 0$. Then, there exists a $\delta_1 > 0$ such that:
$$0 < |x-x_0| < \delta_1 \implies |f(x)-c| < \epsilon$$
$$\implies |f(x)| < |c| + \epsilon$$
Now, let $\delta_2 > 0$ exist such that:
$$0 < |x-x_0| < \delta_2 \implies |g(x)| > M + |c| + \epsilon$
Define $\delta = \min\{\delta_1,\delta_2\}$. Then:
$$0 < |x-x_0| < \delta \implies |f(x)+g(x)| \geq |g(x)| - |f(x)| > M + |c| + \epsilon - |c| - \epsilon = M$$
Since the existence of the desired $\delta$ has been established, this proves the desired result.
Does the proof above work? If not, why? How could I, then, improve it?
This looks correct! One small thing perhaps worth noting is that you didn’t have to do this for all $\epsilon > 0$, but could’ve just chosen $\delta > 0$ such that $0<|x - x_0| < \delta_1 \Rightarrow |f(x) - c| < 7, |g(x)| > M + |c| + 7$, or any other positive number. In fact, $f$ doesn’t even need to have a limit at $x_0$, it just needs to bounded in some neighborhood of $x_0$.