find the limit. use lhospitals rule if appropriate. if there is a more elementary method, co\nlim cot(3x)…

Explanation:

Step1: Rewrite the function

Recall that $\cot(3x)=\frac{\cos(3x)}{\sin(3x)}$, so the function becomes $\lim_{x\rightarrow0^{+}}\frac{\cos(3x)}{\sin(3x)}\cdot\sin(6x)$. Using the double - angle formula $\sin(6x) = 2\sin(3x)\cos(3x)$, we have $\lim_{x\rightarrow0^{+}}\frac{\cos(3x)\cdot2\sin(3x)\cos(3x)}{\sin(3x)}$.

Step2: Simplify the function

Cancel out the $\sin(3x)$ terms in the numerator and denominator. The function simplifies to $\lim_{x\rightarrow0^{+}}2\cos^{2}(3x)$.

Step3: Evaluate the limit

Substitute $x = 0$ into $2\cos^{2}(3x)$. Since $\cos(0)=1$, we get $2\times1^{2}=2$.

Answer:

$2$