Question

L'Hospital's Rule Lim X approaches 0 cot2x•sin6x

Answer 1

$\bf \lim\limits_{x\to 0}\ cot(2x)sin(6x)\implies \lim\limits_{x\to 0}\ \cfrac{cos(2x)}{sin(2x)}sin(6x)\\\\ -----------------------------\\\\ \underline{LH}\qquad \cfrac{-2sin(2x)sin(6x)+cos(2x)6cos(6x)}{2cos(2x)} \\\\\\ \lim\limits_{x\to 0}\ \cfrac{-2sin(2x)sin(6x)+cos(2x)6cos(6x)}{2cos(2x)} \\\\\\ \lim\limits_{x\to 0}\ \cfrac{-0\cdot 0+1\cdot 6\cdot 1}{2\cdot 1}\implies \cfrac{6}{2}\implies 3$