# 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?

• Change the hypothesis to this: $\forall \epsilon>0, \exists \delta>0, \forall x \in D_g, x\in U(x_0, \delta) \rightarrow g(x)>\epsilon$ Jul 15 at 19:57
• Thank you, using this equivalent definition of $lim_{x\to x_0} g(x) = +\infty$ simplifies things a bit. Jul 15 at 20:33

If $$\lim_{x \to x_0}f(x)=A$$, where $$A\in\Bbb{R}$$, then there is a $$\delta>0$$ such that if $$|x-x_0|<\delta$$ then $$|f(x)-A|<1$$. Equivalently, $$A-1.

If $$\lim_{x \to x_0}g(x)=\infty$$, then for every $$N>0$$ there is a $$\delta'>0$$ such that if $$|x-x_0|<\delta'$$, then $$g(x)>N$$.

Combining these inequalities, this means for every $$N>0$$, if $$|x-x_0|<\min(\delta,\delta')$$, then $$f(x)+g(x)>A+N-1$$. Can you finish the proof?

• +1. Or if you like, take $\delta'>0$ s.t. $|x-x_0|<\delta'\implies g(x)>N+|A|+1.$ So if $|x-x_0|<\min (\delta, \delta')$ then $f(x)+g(x)\ge -|A|-1+g(x)>N.$ Jul 18 at 18:04

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:

1. $$A - \dfrac{1}{\epsilon} < f(x) < A + \dfrac{1}{\epsilon}$$
2. $$\dfrac{1}{\dfrac{1}{|A| + \dfrac{2}{\epsilon}}} < g(x)$$ or $$|A|+\dfrac{2}{\epsilon}

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