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#

Related questions
See all questions in Indeterminate Forms and de L'hospital's Rule
Impact of this question
7117 views around the world
You can reuse this answer
Creative Commons License