Limit of Quotient of two functions changed by constant amount

Question

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 $x \to a$.

Answer 1

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.