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Off the top of my head, here are examples that do and don't let you use "limit arithmetic" (addition in both cases).

Ex. 1: $\lim\limits_{x \to \infty}{(\sqrt{x - a} - \sqrt{x})}$

When solving, you cannot do this: $\lim\limits_{x \to \infty}{(\sqrt{x - a} - \sqrt{x})} = \lim\limits_{x \to \infty}{\sqrt{x - a}} - \lim\limits_{x \to \infty}{\sqrt{x}}$

Ex. 2: $\lim\limits_{x \to 0}{\frac{e^{ax}-e^{bx}}{x}}$

Yet, here you can do this: $\lim\limits_{x \to 0}{\frac{e^{ax}-e^{bx}}{x}} = \lim\limits_{x \to 0}{(\frac{e^{ax}-1}{x} - \frac{e^{bx}-1}{x})} = \lim\limits_{x \to 0}{\frac{e^{ax}-1}{x}} - \lim\limits_{x \to 0}{\frac{e^{bx}-1}{x}}$

Why is limit arithmetic is valid in only some cases? When can you use it?

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    $\lim_{x\to a} f(x)g(x)=(\lim_{x\to a} f(x))( \lim_{x\to a} g(x))$ if both limits on RHS exist – 2017-02-23
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    I prefer to call them "algebra of limits" and these rules hold under certain conditions which are always mentioned in the books. Sadly students almost always ignore those conditions when studying these rules. Typical conditions include the existence of limits and nonzero for denominator in handling quotients. By existence I mean the limit is a real number. – 2017-02-23
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    Note further that there are rules to deal with cases when limit is $\pm\infty$ and these complement the algebra of limits. You should see my post http://paramanands.blogspot.com/2013/11/teach-yourself-limits-in-8-hours-part-2.html for more details. – 2017-02-23

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For example, in order to use $$\lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = \lim_{n \to \infty} (a_n+b_n),$$ you need to check that $\lim_{n \to \infty} a_n$ and $\lim_{n \to \infty} b_n$ both exist. Otherwise you might get an indeterminate form and need to resort to things like l'Hôpital's rule.

In your first example, the two limits are $\infty$ so you cannot use the "subtraction arithmetic."

In the second example, you are allowed to do the subtraction because the two limits exist (they are derivatives of $e^{ax}$ and $e^{bx}$ at $x=0$). Note that you could not split it without the clever use of the "$-1$" since $e^{ax}/x\to\infty$ as $x\to 0$.

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    Just to clarify: Is $\infty$ considered indeterminate? – 2017-02-23
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    @SirJony $\infty-\infty$ is indeterminate. To simply go to $\infty$ is not indeterminate--we _know_ what it's doing, it's growing without bound. But if we subtract another thing that goes to $\infty$, we don't know what the difference is doing, that is, whether one of the terms will dominate the other. – 2017-02-23
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    @SirJony To add to David K's comment: note that there are some cases where maybe one or both limits do not exist, yet we can say something about the sum, for example if $a_n \to \infty$ and $b_n \to \infty$, then $a_n+b_n \to \infty$. Or if $a_n \to \infty$ and $b_n \to b$, then $a_n+b_n \to \infty$. – 2017-02-23