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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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    Your step is correct and such transitions are fully justified. See http://math.stackexchange.com/a/1659261/720312016-02-17

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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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    You are, this is mentioned in DonAntonio comment. sorry I forgot to mention that2012-09-28