How do you find lim sin(2theta)/sin(5theta) as theta->0 using l'Hospital's Rule?

Jul 12, 2017

$\setminus {\lim}_{\theta \to 0} \frac{\sin 2 \theta}{\sin 5 \theta} = \frac{2}{5}$

Explanation:

According to L'Hospitol's Rule

$\setminus {\lim}_{\theta \to 0} \frac{f \left(\theta\right)}{g \left(\theta\right)} = \setminus {\lim}_{\theta \to 0} \frac{f ' \left(\theta\right)}{g ' \left(\theta\right)}$, if both $f \left(\theta\right) \to 0$ and $g \left(\theta\right) \to 0$ or both $f \left(\theta\right) \to \infty$ and $g \left(\theta\right) \to \infty$

$\setminus {\lim}_{\theta \to 0} \frac{\sin 2 \theta}{\sin 5 \theta}$

= $\setminus {\lim}_{\theta \to 0} \frac{2 \cos 2 \theta}{5 \cos 5 \theta}$

= $\frac{2 \cos 0}{5 \cos 0} = \frac{2 \times 1}{5 \times 1} = \frac{2}{5}$