Answer
3
Work Step by Step
We know
$$\lim\limits_{\theta \to 0} \dfrac{\mathrm{sin}\left(\theta \right)}{\theta }=1.$$
Thus for any nonzero real number $a$, it follows that
$$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=1.$$
We will use this fact to compute the limit below. So to see that this is true, we let $x=a\theta.$ Then since $x \to 0$ as $\theta \to 0$, we have $$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=\underset{x\to 0}{\lim}\dfrac{\mathrm{sin}\left(x\right)}{x}=1.$$
Now, we want to find
$$\underset{x\to 0}{\lim}[6x^{2}\mathrm{cot}\left(x\right)\mathrm{csc}\left(2x\right)].$$
Note that
$$6{x}^{2}\mathrm{cot}\left(x\right)\mathrm{csc}\left(2x\right)=6{x}^{2}\left(\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}\right)\left(\frac{1}{\mathrm{sin}\left(2x\right)}\right)=2\left(3\right){x}^{2}\left(\frac{1}{\mathrm{sin}\left(x\right)}\right)\left(\frac{1}{\mathrm{sin}\left(2x\right)}\right)\left(\mathrm{cos}\left(x\right)\right)=\left(\frac{x}{\mathrm{sin}\left(x\right)}\right)\left(\frac{2x}{\mathrm{sin}\left(2x\right)}\right)\left(3\mathrm{cos}\left(x\right)\right).$$
Thus
$$\underset{x\to 0}{\lim}[6{x}^{2}\mathrm{cot}\left(x\right)\mathrm{csc}\left(2x\right)]=\underset{x\to 0}{\lim}\left[\left(\frac{x}{\mathrm{sin}\left(x\right)}\right)\left(\frac{2x}{\mathrm{sin}\left(2x\right)}\right)\left(3\mathrm{cos}\left(x\right)\right)\right]\\=\left(1\right)\left(1\right)\left(3\right)=3.$$
