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Since $\lim_{x\to 0} \frac{\sin(x)}{x}=1,\lim_{x\to 0} \frac{x}{\sin(x)}=1 $

$$\lim_{x\to 0} \frac{\cos(x)}{\sin(x)}\cdot\frac{x}{x}=\lim_{x\to 0} \frac{x}{\sin(x)}\cdot\frac{\cos(x)}{x}=\lim_{x\to 0} \frac{x}{\sin(x)}\cdot\lim_{x\to 0}\frac{\cos(x)}{x}=1\cdot\lim_{x\to 0}\frac{\cos(x)}{x}$$

1) I do not understand the transition from this step to this. $$\lim_{x\to 0} \frac{x}{\sin(x)}\cdot\frac{\cos(x)}{x}=\lim_{x\to 0} \frac{x}{\sin(x)}\cdot\lim_{x\to 0}\frac{\cos(x)}{x}$$

I thought that $\lim_{x\to a}A\cdot B= \lim_{x\to a}A\cdot\lim_{x\to a}B$ only if $\lim_{x\to a}A$ and $\lim_{x\to a}B$ exists? In this case, $\lim_{x\to 0}\frac{\cos(x)}{x}$ does not exist. How can we make this transition then?

I use this technique with trigonometric limits very frequently, but I do not understand the basis for this step.

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If this is supposed to be about the behaviour of $\cot x$ near $0$, introducing the $x/x$ seems largely a waste of time. We can (often) make the transition you are talking about. If $\lim_{x\to a}f(x)$ is finite and non-zero, then the limit of $f(x)g(x)$ exists iff the limit of $g(x)$ does. I am not too fond of doing this, since it is too tempting to apply it when it doesn't apply. – André Nicolas Sep 28 '12 at 3:10
I don't know what your teacher required of you, but if I were him/her, I would be much more pleased if my student knew the graph of cotangent and could read the behavior from that. – rschwieb Sep 28 '12 at 11:58
Your step is correct and such transitions are fully justified. See – Paramanand Singh Feb 17 at 4:27
up vote 1 down vote accepted

You are correct.

For example:

$$0=\lim_{x\to\infty}0\cdot \sin(x)\neq \left(\lim_{x\to\infty}0\right)\left(\lim_{x\to\infty}\sin(x)\right)$$

since $\displaystyle{\lim_{x\to\infty}\sin(x)}$ does not exist so there is not meaning to the expression on the LHS.

However, if both limits does exist then it is true that the limit of the product is the product of the limits.

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It isn't necessary that both limits exist: it is enough that one of the functions' limit is zero and that the other function is bounded in some neighbourhood of the point $\,x\,$ approaches for the product to have zero as limit. – DonAntonio Sep 28 '12 at 3:24
I don't get it. So am I allowed to make that step, or not? – Yellow Skies Sep 28 '12 at 3:35
You are, this is mentioned in DonAntonio comment. sorry I forgot to mention that – Belgi Sep 28 '12 at 3:36

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