I know this is a simple question for most of you, but I am currently studying for a Calculus exam and was just wondering why an online calculator I am using to double-check my work was disagreeing with me on this question: $$\lim_{x\to 0} \cot(x)\sec(x)$$
I reduce this down to $\frac{1}{\sin(x)}$, and in that case $x\to 0^-$ the limit is equal to negative infinity; and if $x\to 0^+$, the limit is equal to positive infinity.
Doesn't this mean that the limit as $x\to 0$ does not exist? I use the calculator (linked below), and while it verifies that the two sides approach opposite infinity, it solves the entire limit as approaching "infinity". What does this mean?
Your analysis is correct. Alternatively, $\sec(x)\to 1$ as $x\to 0$, and you can deal with $\cot(x)$, which goes to $\infty$ as $x\to 0^+$ and to $-\infty$ as $x\to 0^-$.
Note, though, the fact that each one-sided limit does not exist is already enough to tell you the limit does not exist. (Saying that the limit equals $\infty$ or $-\infty$ is not saying that the limit exists, it is saying that the limit does not exist and explaining why: because the values of the function grow without bound, either in the positive direction or in the negative direction, respectively). Even though we write things like $$\lim_{x\to 0}\frac{1}{x^2} = \infty$$ this limit does not exist.
As to the limit calculator at your link, I don't know what it means when it says as two-sided limit is $\infty$, since it says the same thing for $\lim\limits_{x\to 0}\frac{1}{x}$. In other words, it means that the on-line calculator is either not giving the correct answer, or else it means something other than what we think it means.