Question

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

Answer 1

`\lim_(theta->0)(sin2theta)/(sin5theta)=2/5`

Explanation:

According to L'Hospitol's Rule

`\lim_(theta->0)(f(theta))/(g(theta))=\lim_(theta->0)(f'(theta))/(g'(theta))`, if both `f(theta)->0` and `g(theta)->0` or both `f(theta)->oo` and `g(theta)->oo`

`\lim_(theta->0)(sin2theta)/(sin5theta)`

= `\lim_(theta->0)(2cos2theta)/(5cos5theta)`

= `(2cos0)/(5cos0)=(2xx1)/(5xx1)=2/5`