Assuming the multiplication property of limits I can do the following:
$\lim \limits_{x \to ∞}f(a)f(b)=\lim \limits_{x \to ∞}f(a)\lim \limits_{x \to ∞}f(b)$
Why cannot do this? The second one is obviously wrong, but I am missing something:
$\lim \limits_{x \to ∞}\frac{n+1}{n+4}=1$
$\lim \limits_{x \to ∞}{n+1}\lim \limits_{x \to ∞}\frac{1}{n+4}=∞*0$
On
You can write
$$\lim_{n\to\infty}(n+1)\frac1{n+4}=\lim_{n\to\infty}\frac{n+1}n\frac n{n+4}=\lim_{n\to\infty}\left(1+\frac1n\right)\frac1{1+\dfrac4n}=\\ \lim_{n\to\infty}\left(1+\frac1n\right)\lim_{n\to\infty}\frac1{1+\dfrac4n}=1\cdot1.$$
Now all terms are defined.
You could also use the rule
$$\lim\frac{f}{g}=\frac{\lim f}{\lim g}$$ which holds provided the limits are defined and $g$ doesn't tend to zero.
The expression $$\lim_{x\to\infty} f(a)f(b)$$
is sloppy, you are sending $x$ to $\infty$, but $x$ doesn't appear later.
Similarly, $$\lim_{x\to\infty}\frac{n+1}{n+4}=1$$ is not true. The limit is equal to $\frac{n+1}{n+4}$.
What you are probably asking is if $$\lim_{x\to\infty} f(x)g(x) = \lim_{x\to\infty}f(x)\lim_{x\to\infty}g(x)$$
and this property is true if both right-hand limits exist.