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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.

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We can set $N=M + c - \epsilon$ at any value and since

$$f(x)+g(x)>M + c - \epsilon=N$$

the proof is complete.

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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.

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Let for every $M>0$ exists $\delta_1$ which satisfies $0 < |x-a| < \delta_1 \implies f(x) > M$.

Good, this is one of the given premises, $\lim_{x\to a}\left(f\left(x\right)\right)=\infty$.

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 $0 < |x-a| < \delta_2 \implies c - \epsilon < g(x) < c + \epsilon$.

Good, this is the other given premise, $\lim _{x\to a}\left(g\left(x\right)\right)=c$ where $c \in R$.

Let for every $N > 0$ exists $\delta$ which satisfies $0 < |x-a| < \delta \implies f(x) + g(x) > N$.

Huh? This is what you have to prove. "Let" sounds like you think you can just assume it is true, like the premises.

But it's good to have written this out somewhere so that you know where you have to go.

So the question is, given an arbitrary $N > 0,$ find an $M$ and $\epsilon$ such that $f(x) > M$ and $|g(x) - c| < \epsilon$ together imply $|f(x) + g(x)| > N$. Although it turns out you can prove $f(x) + g(x) > N$ (as you wrote), which is stronger than the statement with the absolute value.

You have two variables, $M$ and $\epsilon,$ that you can essentially set to whatever you like after you find out what $N$ you have to demonstrate. This gives you a lot of options. You could, for example, just set $\epsilon = 1$ regardless of $N$, and then increase $M$ far enough to make things work.

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You want to prove that for every $M$, there is a $\delta$ such that $|x-a|<\delta\implies f(x)+g(x)>M$.


Take some $\epsilon$ and some $N$ such that $N+c-\epsilon>M$, and by the given two limits you will find $\delta_f$ and $\delta_g$ such that

$$|x-a|<\delta=\min(\delta_f,\delta_g) \implies f(x)>N\land g(x)>c-\epsilon \\\implies f(x)+g(x)>N+c-\epsilon>M.$$

This is because in this neighborhhod of $a$, $f(x)>N$ and $g(x)>c-\epsilon$.

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