Proving a limit using the $\epsilon$ - $\delta$ definition of limit. Given $\lim _{x\to a}\left(f\left(x\right)\right)=\infty$ and $\lim _{x\to a}\left(g\left(x\right)\right)=c$ where $c \in R$, prove $\lim _{x\to a}\left[f\left(x\right)+g\left(x\right)\right]=\infty$.
My attempt:
Let for every $M>0$ exists $\delta_1$ which satisfies $0 < |x-a| < \delta_1 \implies f(x) > M$.
Let for every $\epsilon > 0$ exists $\delta_2$ which satisfies $0 < |x-a| < \delta_2 \implies |g(x) - c| < \epsilon$. Or, I can write it as $0 < |x-a| < \delta_2 \implies c - \epsilon < g(x) < c + \epsilon$.
Let for every $N > 0$ exists $\delta$ which satisfies $0 < |x-a| < \delta \implies f(x) + g(x) > N$.
Using $\delta =$ min{$\delta_1, \delta_2$} so $f(x) > M$ and $g(x) > c - \epsilon$, I get $f(x) + g(x) > M + c - \epsilon$.
And I'm stuck. I've seen a solution somewhere which divides the final equations for $c = 0, c > 0,$ and $c < 0$ but I don't get the idea why do I have to solve it in cases. I don't know how to construct such $\delta$ that satisfies $N$.
I have taken a look at a similar question, proof of limit using epsilon-delta definition, but I'm not quite enlightened with the answer yet.

It's useful to start with stating what you want to prove. In this case:
For every $N>0$ there is $\delta>0$ such that
$$ 0<|x-a|<\delta \Rightarrow f(x) + g(x) > N.$$
So, if you choose your $M$ and $\epsilon$ so that $$ M+c-\epsilon \geq N,$$
you are done.
