Example of danger of limit property I know that if both of the limits
$$
\lim_{x\to a} f(x) \quad\text{and}\quad \lim_{x\to a} g(x)
$$
exist (so they are both equal to real numbers), then
$$
\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x)  + \lim_{x\to a} g(x) 
$$
One could also use the difference or the product instead of the sum. Implied in this is that the limit of the sum of the two functions exist. It is easy to see that the converse of this is not true. That is, the limit of the sum may exist without the limits of the two functions exist.
I can also see how this can be extended to include the cases where, say, $\lim_{x\to a} f(x) = \infty$ (so it doesn't exist) and $\lim_{x\to a} g(x)$ exists (if we understand that $\infty + b = \infty$ for all $b\in \mathbb{R}$.
I see also that people will use this property without first arguing that the two limits exist. For example
$$
\lim_{x\to 1} x + x^2 = \lim_{x\to 1} x + \lim_{x\to 1} x^2 = 1 + 1 = 1
$$
That is, one possibly ought to first state that $\lim_{x\to 1} x$ and $\lim_{x\to 1} x^2$ both exist before using the rule. But all the examples that I have seen of limits, this is never a problem. That is, I don't remember ever seeing a computation of a limit where one uses the rule assume that the two limits exist, but where one of the limits to not in fact exist, but one ends up with an (incorrect) answer.
Finally, here is my question: What is an example of a limit where one falsely uses the rule
$$
\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x)  + \lim_{x\to a} g(x) 
$$
 A: The theorem says if the limits of f(x) and g(x) exist, then the limit of $f(x)+g(x)$ exist and it is equal to the sum of limits.
It does not say anything about the case where limits of $f(x)$ and $g(x)$ do not exist.
Note that $f(x)+g(x)$ is by itself a function and its limit does not necessarily depend on the limit of $f(x)$ or $g(x).$
For example $$\frac {1}{x} + \frac {-1}{x} =0=x+(-x)$$
Where in one case limits do exist and in another case they do not exist but in both cases the limit of the sum does exist.
A: You are correct: the theorem states that if $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist (in $\mathbb R$), then $\lim_{x\to a}f(x)+g(x)$ exists, and $\lim_{x\to a}f(x)+g(x)=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$. The converse of this theorem is not true: in other words, if $\lim_{x\to a}f(x)+g(x)$ exists, then it is not necessarily the case that $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist. For example,
$$
\lim_{x\to 0}\sin\left(\frac{1}{x}\right)+\left(-\sin\left(\frac{1}{x}\right)\right)
$$
exists and equals $0$, but both $\lim_{x\to 0}\sin(1/x)$ and $\lim_{x\to0}-\sin(1/x)$ do not exist. Another natural question is whether it is possible for $\lim_{x\to a}f(x)+g(x)$ exist, given that $\lim_{x\to a}f(x)$ exists, and $\lim_{x\to a}g(x)$ does not exist. The answer is no. Can you see why?
A: If I understand your question correctly, I think that I can show you such an example.
One of my students found $\displaystyle\lim_{x\to\infty}\frac{2x^2+3x+4}{(x+2)(x^2+2)}$ as follows :
$$\begin{align}\lim_{x\to\infty}\frac{2x^2+3x+4}{(x+2)(x^2+2)}&=\lim_{x\to\infty}\bigg(\frac{1}{x+2}+\frac{x+1}{x^2+2}\bigg)
\\\\&=\lim_{x\to\infty}\frac{1}{x+2}+\lim_{x\to\infty}\frac{x+1}{x^2+2}
\\\\&=0+0
\\\\&=0\end{align}$$
and this is correct.
Then, the student tried to find $\displaystyle\lim_{x\to\infty}\frac{-2x^3+2x^2}{(x+2)(x^2+2)}$ as follows :
$$\begin{align}\lim_{x\to\infty}\frac{-2x^3+2x^2}{(x+2)(x^2+2)}&=\lim_{x\to\infty}\bigg(\frac{x^2}{x+2}+\frac{-x^3}{x^2+2}\bigg)
\\\\&=\lim_{x\to\infty}\frac{x^2}{x+2}+\lim_{x\to\infty}\frac{-x^3}{x^2+2}
\\\\&=(+\infty) +(-\infty)\end{align}$$
where the student falsely used the rule
$$
\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x)  + \lim_{x\to a} g(x) 
$$
A: Let me add a couple of more examples.

$$f(x)=\csc x,\quad g(x)=-\cot x$$
$$\lim_{x\to0}\left(f(x)+g(x)\right)=\lim_{x\to0}\tan \frac x2=0$$
but, either $\lim\limits_{x\to0}f(x)$ or $\lim\limits_{x\to0}g(x)$ doesn't exist.

$$f(x)=\lfloor x\rfloor,\quad g(x)=\lfloor -x\rfloor$$
$$\lim_{x\to0}\left(f(x)+g(x)\right)=-1$$
but, either $\lim\limits_{x\to0}f(x)$ or $\lim\limits_{x\to0}g(x)$ doesn't exist.
