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Let $f(x), g(x)$ be two function's, how to show that $$ \limsup_{x\to a} \frac{f(x)}{g(x)} = \limsup_{x\to a} \frac{f(x)}{g(x) + c} $$ for $a \in \mathbb R \cup \{ \pm \infty \}$ and $c \in \mathbb R$.

Edit From the comments below: $f$ and $g$ are both unbounded as $x \to a$.

Arthur
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StefanH
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    I think you need some condition on the behaviour of $f$ and $g$ as $x \to a$, otherwise this is false. Consider, for example, the functions $f(x) = g(x) = 1$. – Arthur Aug 09 '15 at 18:32
  • Sorry, both $f(x), g(x)$ are unbounded for $x \to a$, i.e. $f(x), g(x) \to \infty$ for $x \to a$. – StefanH Aug 09 '15 at 18:37
  • have you checked what happens when you put $\limsup_{x\to a} \frac{f(x)}{g(x) + c}=\limsup_{x\to a} \frac{f(x)}{g(x)}\frac 1 {1+c/g(x)}$? – user190080 Aug 09 '15 at 18:42
  • Guess then it is solved: $$ \limsup_{x\to a} \frac{f(x)}{g(x)} \frac{1}{1+c/g(x)} = \limsup_{x\to a} \frac{f(x)}{g(x)}\cdot \limsup_{x\to a} \frac{1}{1+c/g(x))} = \limsup_{x\to a} \frac{f(x)}{g(x)} \cdot 1. $$ – StefanH Aug 09 '15 at 18:50
  • the problem is, that this multiplication equality is not true in general for the limit superior...so one definitely needs to argue in this step a bit more – user190080 Aug 09 '15 at 19:22
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    Unboundedness of $g$ does not mean that you always have $g(x) \to \infty$. I think the latter condition is what you actually want to deal with, otherwise we have a counterexample. – Sangchul Lee Aug 09 '15 at 19:24

1 Answers1

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Following this, we want to check that for unbounded functions such that for $x\to a,g(x),f(x)\to \pm \infty$ it holds $$ \limsup_{x\to a} \frac{f(x)}{g(x)} = \limsup_{x\to a} \frac{f(x)}{g(x) + c} $$ It seems that we need to assume further, that both $f,g$ either go to plus infinity or both go to minus infinity at the same time. So that $\frac f g$ stays positive in a surrounding of $a$. If $\limsup_{x\to a} \frac{f(x)}{g(x)}=\infty$ the reasoning which follows should still hold.

First we do a simple modification $$ \limsup_{x\to a} \frac{f(x)}{g(x) + c}=\limsup_{x\to a}\left( \frac{f(x)}{g(x)}\frac 1 {1+\frac c {g(x)}}\right ) $$ which is certainly true since $g(x)\neq 0$ in a neighborhood of $a$ because of our unbounded assumption.

Then we conclude, using the property of $\limsup$ and some limiting point $x_0$ $$ \limsup_{x\to x_0}\left(f(x)g(x)\right )\leq \limsup_{x\to x_0}f(x)\limsup_{x\to x_0}g(x) $$ that indeed it holds

\begin{align} \limsup_{x\to a} \frac{f(x)}{g(x) + c}=\limsup_{x\to a}\left( \frac{f(x)}{g(x)}\frac 1 {1+\frac c {g(x)}}\right )&\leq\limsup_{x\to a} \frac{f(x)}{g(x)}\limsup_{x\to a}\frac 1 {1+\frac c {g(x)}}\tag 1 \\ &=\limsup_{x\to a}\frac{f(x)}{g(x)} \end{align} because $$ \limsup_{x\to a}\frac 1 {1+\frac c {g(x)}}=\liminf_{x\to a}\frac 1 {1+\frac c {g(x)}}=\lim_{x\to a}\frac 1 {1+\frac c {g(x)}}=1 $$ The other direction also holds, since \begin{align} \limsup_{x\to a} \frac{f(x)}{g(x)}=\limsup_{x\to a} \frac{f(x)}{g(x)}\liminf_{x\to a}\frac 1 {1+\frac c {g(x)}}\leq&\limsup_{x\to a}\left( \frac{f(x)}{g(x)}\frac 1 {1+\frac c {g(x)}}\right ) \tag 2\\=&\limsup_{x\to a} \frac{f(x)}{g(x) + c} \end{align} Eventually because of the inequalities $(1)$ and $(2)$ the equality holds.

For further references one could check the linked references in the first given link from above. The case where $f/g$ is negative needs to get some further attention - it's possible that there exist counterexamples.

Community
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user190080
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  • Do you know an example when $\limsup f(x)g(x) < \limsup f(x) \cdot \limsup g(x)$ for $\limsup f(x) < \infty, \limsup g(x) < \infty$? – StefanH Aug 12 '15 at 15:45
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    @Stefan yeah, for example let's say $f(x)=1$ for $x$ rational and otherwise $f(x)=2$, $g(x)=1/2$ for $x$ irrational and $g(x)=1$ otherwise, so the product $f(x)g(x)=1$ for all $x\in\mathbb{R}$ Therefore $\limsup_{x\to x_0}f(x)g(x)=1$ but $\limsup_{x\to x_0}f(x)=2$ and $\limsup_{x\to x_0}g(x)=1$ and therefore we have $\limsup_{x\to x_0}f(x)g(x)=1<\limsup_{x\to x_0}f(x)\limsup_{x\to x_0}g(x)=2$ for $x_0\in\mathbb{R}$ – user190080 Aug 12 '15 at 21:10