If $\lim_{x\to a} |g(x)| =\lim_{x\to a} |\frac{f(x)}{g(x)}| = \infty$, is it true that $\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x)$? Let $\lim_{x\to a} |g(x)| =\lim_{x\to a} |\frac{f(x)}{g(x)}| = \infty$ for some $a\in \mathbb R$. I would like to show that $\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x)$. Below is my attempt.
Without losing generality, let us assume that $\lim_{x\to a} \frac{f(x)}{g(x)} = \infty$. It can be seen that $\infty = \lim_{x\to a} \frac{f(x)}{g(x)} + 1 = \lim_{x\to a} \frac{f(x)+g(x)}{g(x)}$. Therefore, $\lim_{x\to a} f(x) + g(x) = (\lim_{x\to a} \frac{f(x)+g(x)}{g(x)}) (\lim_{x\to a} g(x)) = (\lim_{x\to a} \frac{f(x)}{g(x)})(\lim_{x\to a} g(x)) = \lim_{x\to a} f(x).$
Whereas the last line follows from the "product of limits equal limit of product" rule for the $\pm$infinity/$\pm$infinity cases.
Is the proof good enough to demonstrated the result? If not, is this true in the first place? It would be appreciated to receive some help from you.
 A: I think this is true. As Greg Martin says, it's only reliable to prove it from the definitions. Throughout the proof I'll only write one $\delta >0$, because we can take the minimum of all $\delta_i$'s and call it $\delta$. \
Let $M>1$ there exists $\delta >0$ such that $|\frac{f(x)}{g(x)}|>\sqrt{M}$  and $|g(x)|>\sqrt{M}$ whenever $0<|x-a|<\delta$, this implies $|f(x)|>\sqrt{M}|g(x)|>M$ whenever $0<|x-a|<\delta$ (note that same is true for any $M>0$). So $\lim_{x \rightarrow a}|f(x)|=\infty$. To prove it we must assume the existence of $\lim_{x \rightarrow a}f(x)$. Now we can have the following cases:\
(i)     Let $\lim_{x \rightarrow a}f(x)=\infty$ and $\lim_{x \rightarrow a}g(x)=-\infty$. Choose $M_1>0$ there exists $\delta >0$ such that $f(x)>M_1>0$ and $g(x)< -\frac{M_1}{\sqrt{M}-1}<0$ whenever $0<|x-a|<\delta$. Then for $0<|x-a|<\delta$ , $|f(x)|=f(x)$ and $|g(x)|=-g(x)$. Let $0<|x-a|<\delta$ then:
\begin{align}
&|f(x)|>\sqrt{M}|g(x)|=(\sqrt{M}-1)|g(x)|+|g(x)| \\
&\Rightarrow f(x)-|g(x)| > (\sqrt{M}-1)|g(x)|\\ 
&\Rightarrow f(x)+g(x)> (\sqrt{M}-1)(-g(x))>M_1
\end{align}
Thus $\lim_{x \rightarrow a} (f(x)+g(x))=\infty=\lim_{x \rightarrow a}f(x)$. \
(ii)     Let $\lim_{x \rightarrow a}f(x)=-\infty$ and $\lim_{x \rightarrow a}g(x)=\infty$. Choose $M_1>0$ there exists $\delta >0$ such that $f(x)<-\frac{M_1}{1-\frac{1}{\sqrt{M}}}<0$ and $g(x)> M_1>0$ whenever $0<|x-a|<\delta$. Then for $0<|x-a|<\delta$ , $|f(x)|=-f(x)$ and $|g(x)|=g(x)$. Let $0<|x-a|<\delta$ then:
\begin{align}
&|f(x)|>\sqrt{M}|g(x)| \\
&\Rightarrow -f(x)>\sqrt{M}g(x)\\
&\Rightarrow -\frac{1}{\sqrt{M}}f(x)>g(x)\\
&\Rightarrow (1-\frac{1}{\sqrt{M}})f(x)-f(x)>g(x)\\
&\Rightarrow f(x)+g(x) < (1-\frac{1}{\sqrt{M}})f(x)<-M_1
\end{align}
Thus $\lim_{x \rightarrow a} (f(x)+g(x))=-\infty=\lim_{x \rightarrow a}f(x)$.\
(iii)      Let $\lim_{x \rightarrow a}f(x)=-\infty$ and $\lim_{x \rightarrow a}g(x)=-\infty$. Choose $M_1>0$ there exists $\delta >0$ such that $f(x)<-\frac{M_1}{2}$ and $g(x)<- \frac{M_1}{2}$ whenever $0<|x-a|<\delta$. Then for, $0<|x-a|<\delta$, $f(x)+g(x)<-M_1$
Thus $\lim_{x \rightarrow a} (f(x)+g(x))=-\infty=\lim_{x \rightarrow a}f(x)$.\
(iv)   The case $\lim_{x \rightarrow a}f(x)=\infty$ and $\lim_{x \rightarrow a}g(x)=\infty$ can be done in the same way as (iii).
(v)   Let $\lim_{x \rightarrow a}f(x)=\infty$, $\lim_{x \rightarrow a-}g(x)=-\infty$ and $\lim_{x \rightarrow a+}g(x)=\infty$ (the opposite  limits of $g$ will be similar). Then from (i) considering the neighborhood $(a-\delta, a)$ we get $\lim_{x \rightarrow a-} (f(x)+g(x))=\infty=\lim_{x \rightarrow a-}f(x)=\lim_{x \rightarrow a}f(x)$. Then from (iv) considering the neighborhood $(a,a+\delta)$ we get
$\lim_{x \rightarrow a+} (f(x)+g(x))=\infty=\lim_{x \rightarrow a+}f(x)=\lim_{x \rightarrow a}f(x)$. Combining those we have $\lim_{x \rightarrow a-} (f(x)+g(x))=\lim_{x \rightarrow a}f(x)=\lim_{x \rightarrow a+} (f(x)+g(x))$ which implies
$\lim_{x \rightarrow a} (f(x)+g(x))=\lim_{x \rightarrow a}f(x)$.
(vi) The case when $\lim_{x \rightarrow a}f(x)=-\infty$, $\lim_{x \rightarrow a-}g(x)=-\infty$ and $\lim_{x \rightarrow a+}g(x)=\infty$ (the opposite  limits of $g$ will be similar) is done similayly.
