Answer
The mentioned statement is false.
Work Step by Step
$\lim\limits_{x \to 4}\Big(\frac{2x}{x-4}-\frac{8}{x-4}\Big)=\lim\limits_{x \to 4}\frac{2x}{x-4}-\lim\limits_{x \to 4}\frac{8}{x-4}$
We see that as $x\to 4$, $2x\to8$ and $(x-4)\to0$
So, $(\frac{2x}{x-4})\to+\infty$ and $(\frac{8}{x-4})\to+\infty$
Which means, both $\lim\limits_{x \to 4}\frac{2x}{x-4}$ and $\lim\limits_{x \to 4}\frac{8}{x-4}$ do not exist.
However, the difference law $$\lim\limits_{x \to a}[f(x)-g(x)]=\lim\limits_{x \to a}f(x)-\lim\limits_{x \to a}g(x)$$can only be applied on the assumption that both $\lim\limits_{x \to a} f(x)$ and $\lim\limits_{x \to a}g(x)$ exist.
Therefore, the mentioned statement is false.
