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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:

$\lim \limits_{x \to ∞}\frac{n+1}{n+4}=1$

$\lim \limits_{x \to ∞}{n+1}\lim \limits_{x \to ∞}\frac{1}{n+4}=∞*0$

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    $\infty \times 0$ is undefined. The multiplication property cannot be used if either limit fails to exist. – 2017-01-27
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    As $x$ approaches $\infty$, the value of $n$ is unaffected...I think you have some typos. – 2017-01-27
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    You can't do it because it gives you nonsensical answers, precisely in examples such as the one you present. – 2017-01-27
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    You "can" do it, and it leads you nowhere as this product is undefined. – 2017-01-27
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    If both limits are finite and nonzero its okay to do. Also, yea, you have a terrible use of $\lim$ and using different variables than what your limit is supposedly moving. $\lim\limits_{x\to\infty}f(x)g(x)$ is what you mean to write, and $\lim\limits_{x\to\infty}\frac{x+1}{x+4}$. Further, the tag (education) is completely irrelevant to your question. – 2017-01-27

2 Answers 2

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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.

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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.