How to prove $\lim_{x\to x_0}\left(f(x)+g(x)\right)=+\infty$ if $f(x)\rightarrow A$ and $g(x)\rightarrow +\infty$? I want to prove - if $\lim_{x\to x_0} f(x) = A \in \mathbb{R}$ and $\lim_{x\to x_0} g(x) = +\infty$ then $\lim_{x \to x_0} \left(f(x) + g(x)\right) = +\infty$ (where $x_0 \in \bar{\mathbb{R}}$), using Cauchy definition.
Translating it into $\epsilon,\delta$ language it becomes:
Hypothesis 1. $\forall \epsilon>0, \exists \delta>0, \forall x \in D_f, x\in U(x_0, \delta) \rightarrow A-\epsilon < f(x) < A + \epsilon$.
Hypothesis 2. $\forall \epsilon>0, \exists \delta>0, \forall x \in D_g, x\in U(x_0, \delta) \rightarrow \frac{1}{\epsilon} < g(x)$
Goal. $\forall \epsilon>0, \exists\delta>0, \forall x\in D_f \cap D_g, x\in U(x_0, \delta) \rightarrow \frac{1}{\epsilon} < f(x) + g(x)$
So, I have arbitrary $\epsilon$, choose some $\epsilon_1>0$ and $\epsilon_2>0$ to put into both hypotheses, get corresponding $\delta_1>0$ and $\delta_2>0$... then I use value $\min(\delta_1, \delta_2)$... skipping forward, I get to the following: I have to prove from $A-\epsilon_1 < f(x) < A + \epsilon_1$ and $\frac{1}{\epsilon_2} < g(x)$ that $\frac{1}{\epsilon} < f(x) + g(x)$ follows.
How to choose $\epsilon_1$ and $\epsilon_2$ to do that? If $A > 0$ it seems I can choose $\epsilon_1 = A$ and $\epsilon_2 = \epsilon$, but what to do when that's not case?

Anyway, answer to my question (with the $1/\epsilon$ kind of definition of positive infinity limit):
$\epsilon_1 = \dfrac{1}{\epsilon}$ and $\epsilon_2 = \dfrac{1}{|A| + \dfrac{2}{\epsilon}}$.
That results in inequalities:

*

*$A - \dfrac{1}{\epsilon} < f(x) < A + \dfrac{1}{\epsilon}$

*$\dfrac{1}{\dfrac{1}{|A| + \dfrac{2}{\epsilon}}} < g(x)$ or $|A|+\dfrac{2}{\epsilon}<g(x)$
Summing them, we get $A+|A|+\dfrac{1}{\epsilon} < f(x) + g(x)$ and $\dfrac{1}{\epsilon} < f(x) + g(x)$ follows.
