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`