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$\lim\limits_{x \to 0} f(x) = \dfrac{\sin2x}{x\cos3x}$

By the product law, can't we write:

$= 2 \cdot \lim\limits_{x \to 0} \dfrac{\sin2x}{2x} \cdot \lim\limits_{x \to 0} \dfrac{1}{\cos3x}$

Then taking the limits, replace $2x$ with $\theta$ and $\theta$ approaches $0$ if you like

$= 2 \cdot 1 \cdot \dfrac{1}{1}$

$= 2$ ?

However wolfram says it is $-\infty$ on one side and $+\infty$ on the other, and I am inclined to believe it :)

Where am I messing up?

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    @DavidMitra, my apologies, I did not bracket cos(3x), I wrote cos3x, and you are right, my input was wrong. Perhaps somebody can delete this question for me.2012-09-16

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You are correct. It appears you entered the limit into Wolfram|Alpha incorrectly, as it gives me $2$ as well here.

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    Thanks Alex, I see now that cos(3x) is necessary, cos3x is interpreted differently than I expected, I should have noticed in the nicely formatted equation wolfram presents!2012-09-16