Im trying to prove If $\lim_{x \rightarrow a}f(x) = 100$ and $\lim_{x \rightarrow a} g(x) = \infty$ then $\lim_{x \rightarrow a} (f(x) + g(x)) = \infty$ using epsilon-delta
$\lim_{x \rightarrow a} g(x) = \infty \iff \forall C \: \exists\delta_1 > 0, \: 0<|x-a| <\delta_1 \implies g(x) > C $
$\lim_{x \rightarrow a} f(x) = 100 \iff \forall \epsilon >0 \: \exists \delta_2 > 0, \: 0<|x-a| <\delta_2 \implies |f(x) - 100| < \epsilon $
I'll follow up from my comments with a more complete hint at a solution.
Before we delve into the formal proof it's important to see intuitively why the statement is true. The point is that eventually we can be sure $f(x) > 0$ (because $100 > 0$). Then adding it to $g(x)$ only gets us a bigger number, so we're certainly still tending to infinity!
As you pointed out, to show that $\lim_{x\to a}(f(x) + g(x)) = \infty$ we need to prove the statement:
For all $M$ there is some $\delta >0$ so that if $|x-a| < \delta$ then $f(x) + g(x) > M$.
Recall our intuition. We want to get to a point where $f$ is positive. Luckily we can do that! Pick $\delta_1$ so that if $|x-a| < \delta_1$ then $|f(x) - 100| < 100$. In other words this means that if $|x-a| < \delta_1$ then $0 < f(x) < 200$.
We also want $g$ to get large enough that it becomes bigger than $M$. We can also do that. Since $g(x) \to \infty$ as $x \to a$, there is a $\delta_2$ so that if $|x-a| < \delta_2$ then $g(x) > M$.
Now, see if you can put these results together to finish the proof.
The key takeaway is this: it's perfectly fine to plug values in to the epsilon-delta definitions! In fact, it makes proofs much easier.