1

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.

Minh Khôi
  • 963
  • I'm more concerned with the case in which both $f$ and $g$ have infinite limits but different signs. – Minh Khôi Mar 01 '21 at 02:56
  • 1
    $f(x)+g(x)\to f(x)$ is not a well-defined mathematical statement. (You also haven't explicitly defined what you mean by $\ll$.) It sounds like you're trying to prove something like: if $\lim_{x\to a} f(x)=\infty$ and $\lim_{x\to a} g(x)=-\infty$ and $\lim_{x\to a} \frac{f(x)}{g(x)}=-\infty$, then $\lim_{x\to a} \big( f(x)+g(x) \big)=\infty$ as well; is that accurate? – Greg Martin Mar 01 '21 at 04:55
  • @GregMartin: That is accurate. – Minh Khôi Mar 01 '21 at 05:14
  • 1
    In that case, I recommend first proving that $f(x)+g(x) > \frac{f(x)}2$ when $x$ is near $a$, and then use a comparison test. More generally, trying to apply limit laws that work for finite limits is not a rigorous or reliable way to prove limit laws for infinite limits; it's usually better to go back to the definitions of infinite limits and prove what you want directly. – Greg Martin Mar 01 '21 at 06:21
  • Thank you for your advice. Rigorous proof is always better. But I couldn't derive a proper one. Besides, isn't for the case of the product law we get the immediate result? Since $\infty \times \infty$ is not an indeterminate form. I would also like to know if my proof contains any err in it. – Minh Khôi Mar 01 '21 at 11:02
  • 1
    What if $\lim_{x \rightarrow a} g(x)$ does not exists? Take $a=0,g(x)=\frac{1}{x}$ – absolute0 Mar 01 '21 at 13:46
  • I see, thank you very much for pointing that out. – Minh Khôi Mar 02 '21 at 03:37

1 Answers1

1

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.

absolute0
  • 1,027