Question

Solve $\underset{x\to 0}{lim}\frac{\sin 4x}{\sin 2x}$

Answer 1

$\underset{x\to 0}{lim}\frac{sin4x}{sin2x}$

Apply L Hopital's Rule

$=\underset{x\to 0}{lim}\frac{cos\left(4x\right).4}{cos\left(2x\right).2}$

$=\underset{x\to 0}{lim}\frac{2cos\left(4x\right)}{cos\left(2x\right)}$

$=\frac{2cos\left(4\right)}{cos\left(2\right)}$

$=2$