Although closely related:
i) $\lim_{x\rightarrow a}f(x) = L; a\in \mathbb R$
ii) $\lim_{x\to a^+}f(x) = L$
iii) $\lim_{x\to a^-}f(x) = L$
iv) $\lim_{x\to \infty} f(x) = L$
and v) $\lim_{x\to -\infty} f(x) = L$.
have technically different definitions.
i) means as $x$ gets close to $a$, then $f(x)$ gets close to $L$. or technically:
i) means For any $\epsilon > 0$ we can find $\delta$ so that whenever $|x -a| < \delta$ it will follow that $|f(x) - L| < \epsilon$.
ii) means as $x$ gets close to $a$ but larger than $a$, then $f(x)$ gets close to $L$. or technically:
ii) means For any $\epsilon > 0$ we can find $\delta$ so that whenever $a < x < a+ \delta$ it will follow that $|f(x) - L| < \epsilon$.
Notice that if i) is true than ii) must be true but if ii) is true i) is not nescessarily true.
iii) means as $x$ gets close to $a$ but smaller than $a$, then $f(x)$ gets close to $L$. or technically:
iii) means For any $\epsilon > 0$ we can find $\delta$ so that whenever $a -\delta < x < a$ it will follow that $|f(x) - L| < \epsilon$.
Notice that if i) is true than ii) and iii) must be true. And if both ii) and iii) is true then i) is true. But if one or the other of ii) or iii) is true but the other isn't i) will not be true.
So the statement "i) if ii) and iii)" is not a definition but an observation. Well.... it could be a definition as they are equivalent.
Now...
Limit to infinity is not that as $x$ gets close to infinity $f(x)$ will get close to $L$. That wouldn't make any sense because we can't get $x$ "close to infinity". Instead we mean, as $x$ becomes large, $f(x)$ gets close to $L$. or technically:
iv) For any $\epsilon > 0$ there is some $M \in \mathbb R$ so that whenever $x > M$ it will follow that $|f(x) - L| < \epsilon$.
It makes no sense to talk of $\lim_{x\to \infty^+} f(x)$ because we can't "approach from the right". And to talk of $\lim_{x \to \infty^-}f(x)$ is the exact same thing as $\lim_{x \to \infty}f(x)$ because we can only "approach from the left".
So it is NOT the case that $\lim_{x\to\infty} f(x) = L$ if $\lim_{x\to \infty^+}f(x) = L$ and $\lim_{x\to \infty^-}f(x) = L$. Instead it is just $\lim_{x\to\infty} f(x) = L$ if $\lim_{x\to \infty^-}f(x) = L$.
$\lim_{x \to -\infty}f(x)=L$ means we are talking the values of very big negative numbers (very small, or negative with very large magnitude or absolute values). It should be very clear that $-\infty$ means infinity beyond the negative extreme of the reals, whereas $\infty^-$ means something entirely different; it means approaching the the positive extreme of the real numbers but at some finite value less than infinity.
v) For any $\epsilon > 0$ there is some $M \in \mathbb R$ so that whenever $x < M$ it will follow that $|f(x) - L| < \epsilon$.
Again it makes no sense to talk of $\lim_{x\to -\infty^-}f(x)$ and it is redundant to talk of $\lim_{x\to -\infty^+} f(x)$ for the exact same reasons.
Best Answer
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.