# Limit Proof

1. Jul 1, 2014

### Tsunoyukami

Suppose $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = c$ where $c$ is a real number. Prove $\lim_{x \to a} \big( f(x)+g(x) \big) = \infty$.

Proof Assume $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = c$. Then, by definition:

1) For every $M > 0$ there exists $\delta_{1} > 0$ such that $0<|x-a|<\delta_1 \implies f(x)>M$.
2) For every $\epsilon >0$ there exists $\delta_{2} >0$ such that $0<|x-a|<\delta_{2} \implies |g(x)-c|<\epsilon$.

We want to show that for every $S>0$ there exists $\delta>0$ such that $0<|x-a|<\delta \implies f(x)+g(x)>S$.

Choose $\delta = \min(\delta_{1}, \delta_{2})$.
Then $0<|x-a|<\delta \implies f(x)>M$ and Then $0<|x-a|<\delta \implies |g(x)+c|<\epsilon \implies c -\epsilon < g(x) < c + \epsilon$.

Then $f(x) + g(x) > M - \epsilon + c = S$ and so, with $S = M + c - \epsilon$ we conclude $\lim_{x \to a} \big( f(x)+g(x) \big) = \infty$.

Is this a valid proof? I'm not entirely sure if my argument is valid near the end. Any help is appreciated! Thanks.

2. Jul 1, 2014

Seems legit.

3. Jul 2, 2014

### Fredrik

Staff Emeritus
It's good, but there are a few things you can improve. g(x)+c appears to be a typo. Should be g(x)-c. I think it's not entirely clear what you're doing with S and M. It would add some clarity if you say "let S>0 be arbitrary" at the start of the proof, and then make it clear that you have chosen M and ε to be positive real numbers such that M-ε+c≥S (or =S). (It kind of looks like you let M and ε be arbitrary and then defined S by S=M-ε+c. That's not OK). That last line needs a "for all x such that |x-a|<δ" or a $|x-a|<\delta~\Rightarrow~$.

Last edited: Jul 2, 2014