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I am trying to prove: If $\lim_{x\to a^+}f(x)=\infty$ and $\lim_{x\to a^+}g(x)=\infty$, then $\lim_{x\to a^+}(f(x)+g(x))=\infty$

This is what I have:

Proof:

$\forall M_1>0$, $\exists\delta_1>0:0<x-a<\delta_1\implies f(x)>M_1$.

$\forall M_2>0$, $\exists\delta_2>0:0<x-a<\delta_2\implies g(x)>M_2$.

Let $M>0$. Choose $\delta=min\{\delta_1,\delta_2\}.$

Thus $\delta\leq \delta_1$ and $\delta\leq\delta_2$.

Assume $0<x-a<\delta\leq\delta_1,\delta_2$.

So, $f(x)>M_1$ and $g(x)>M_2$.

Show $f(x)+g(x)>M$.

I get stuck here and I don't know what to do. If anybody could help, that would be greatly appreciated.

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Take $M>0$. Since $\frac M2>0$, there is some $\delta_1>0$ such that $a<x<a+\delta_1\implies f(x)>\frac M2$. And there is some $\delta_2>0$ such that $a<x<a+\delta_2\implies g(x)>\frac M2$. So, if $\delta=\min\{\delta_1,\delta_2\}$,$$a<x<a+\delta\implies f(x)+g(x)>M.$$

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