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

1 Answer
Shwetank Mauria
Jul 12, 2017

\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

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