# Manipulating trig limit functions

I'm having difficulty understanding how my calc teacher manipulated this problem

What I dont understand is how he minipulated the second line to the third line. In other words, I don't understand how he changed $6x^2$ to $6\cos3x$. And how he changed $\displaystyle \frac{1}{\sin2x}$ to $\displaystyle \frac{x}{\sin2x}$. Unfortunately I have no way of getting in contact with him and the test is tomorrow. Any help would be greatly appreciated.

sorry about the external URL, I just don't know of any quicker way to add the math to this post.

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He didn't change anything. Note that $x^{2}=x \cdot x$ so that by changing $6x^{2} \cos 3x$ to $6 \cos 3x$, that "frees up" the $x^{2}$ to make the other changes such as changing $\frac{1}{\sin 2x}$ to $\frac{x}{\sin 2x}$.

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That's not what your teacher did. He rewrote the entire expression:

\begin{align} 6x^2 \cot 3x \csc 2x &= 6x^2\cdot \frac{\cos 3x}{\sin 3x} \cdot \frac{1}{\sin 2x}\\ &= 6\cos 3x \cdot \frac{x}{\sin 3x} \cdot \frac{x}{\sin 2x} \end{align}

and then he took the limit of each factor separately.

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Hint:

Utilize following two identities :

$$\lim_{y\to0}\frac{\sin y}y=1\text{ and } \lim_{y\to0}\cos y=1$$

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