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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?

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It doesn't mean anything special. The limit doesn't exist. The one sided limits (if you allow these symbols) are $+\infty$ and $-\infty$. So the 'calculator' is actually wrong. I wouldn't rely on it. – Weltschmerz Feb 25 '11 at 3:58
up vote 6 down vote accepted

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.

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