Epsilon - Delta proof of limit properties at infinity.

Question

I want to show some basic properties of limits tending to infinity in point $a \in A$ with $f: A \subseteq \mathbb{R} \to \mathbb{R} $ with just the defenitions of limits. I want to show following properties:

$1. \ (+ \infty) + (+ \infty) = + \infty$

$2. \ for \ every \ L \in \mathbb{R} \ is: L + (- \infty) = - \infty$

$3. \ for \ every \ L \in \mathbb{R} \ is: \frac{L}{+ \infty} = 0 $

My attempt:

Property 1:

I tried it with the following argument. If you have two functions $f$ and $g$ which tend to infinity, than for $f$ we can find a $\delta_1 > 0$ for every $M \in \mathbb{R}$ so that $0 < |x-a| < \delta_1 \implies f(x) > M$ . We also can find for $g$ a $\delta_2$ and $N$ with the same properties. So if we take $\delta = min\{\delta_1,\delta_2 \}$ than we can add $f$ and $g$ so we get for $0 < |x-a| < \delta \implies f(x) + g(x) > M + N$. This implies that the sum of the functions tends to infinity in $a$. So we can say that $ \ (+ \infty) + (+ \infty) = + \infty$

Property 2:

I think that this proof is similar to property 1, but I don't know how to use the definition of a normal limit: $\lim_{x\to a} f(x) = L$ if only if $\forall \epsilon> 0, \exists \delta > 0, \forall x \in A :0 < |x-a| < \delta \implies |f(x) - L| < \epsilon$

Property 3:

I don't know where to start here.

Can someone give me a full proof of these properties?

Thanks in advance

Answer 1

1. It's fine.
2. Take $M<0$. There's a $\delta_1>0$ such that $|x-a|<\delta_1\implies f(x)<M-L-1$. And there is a $\delta_2>0$ such that $|x-a|<\delta_2\implies\bigl|g(x)-L\bigr|<1\implies g(x)<L+1$. Therefore, if $\delta=\min\{\delta_1,\delta_2\}$, then $|x-a|<\delta\implies f(x)+g(x)<M$.
3. Take $\varepsilon>0$. There's a $\delta_1>0$ such that$$|x-a|<\delta_1\implies\bigl|g(x)-L\bigr|<1\implies\bigl|g(x)\bigr|<L+1.$$And there is a $\delta_2>0$ such that $|x-a|<\delta\implies f(x)>\frac{L+1}\varepsilon$. So, if $\delta=\min\{\delta_1,\delta_2\}$, then $|x-a|<\delta\implies \frac{g(x)}{f(x)}<\varepsilon$.