What is the value of $\frac{\lim_{x\to 0^+}f(x)}{\lim_{x\to 0^+}g(x)}$

Let, $A=\mathbb R-\left\{0\right\}$ with

$$f:A\rightarrow\mathbb R,\quad g:A\rightarrow\mathbb R$$

$g(x)=\frac{2x}{|x|}$

$f(x)=\begin{cases}8-4x-g(-x),\,x≥0\\g(x)+x+6,\,x<0\end{cases}$

Then, what is the value of

$$\frac{\lim_{x\to 0^+}f(x)}{\lim_{x\to 0^+}g(x)}=?$$

My attempt.

$$\lim_{x\to 0^+}g(x)=\frac{2x}{x}=2$$

$$\lim_{x\to 0^+}f(x)=8-\lim_{x\to 0^+}g(-x)=8-\lim_{x\to 0^+}\frac{-2x}{x}=8+2=10$$

So, $$\frac{\lim_{x\to 0^+}f(x)}{\lim_{x\to 0^+}g(x)}=5$$

My friend says me, my answer is correct, but my attempt is wrong.

She says me

$$\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\left(8-\color{red}{\frac{2x}{-x}}\right)=8+2=10$$

But, I can not understand my mistake.

Nothing is wrong with your attempt. My guess is that your friend evaluated $\lim_{x\to 0^+} g(-x)$ differently. I can see two obvious ways to do it:

1. $\lim_{x\to 0^+} g(-x) = \lim_{x\to 0^+} \dfrac{2(-x)}{|-x|} = \lim_{x\to 0^+}\dfrac{-2x}{x} = -2,$
2. $\lim_{x\to 0^+} g(-x) = \lim_{t\to 0^-} g(t) = \lim_{t\to 0^-} \dfrac{2t}{|t|} = \lim_{t\to 0^-}\dfrac{2t}{-t} = -2.$

Since you calculated $\lim_{x\to 0^+} g(x)$ beforehand, you could also just notice that $g$ is an odd function, i.e. $g(-x) = -g(x)$. Then,

3. $\lim_{x\to 0^+} g(-x) = \lim_{x\to 0^+} (-g(x)) = - \lim_{x\to 0^+} g(x)$.