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

Question

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

Paisley Johnson · Accepted Answer

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

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#

Scarlett Allison · Answer

undefined To find $ \lim_{\theta \to 0} \frac{\sin(2\theta)}{\sin(5\theta)} $ using l'Hospital's Rule:

1. Apply l'Hospital's Rule:
   - Differentiate both the numerator and the denominator with respect to $ \theta $.
   - Compute the limit of the resulting expression as $ \theta $ approaches 0.

2. Differentiate the numerator:
   - $ \frac{d}{d\theta} \sin(2\theta) = 2\cos(2\theta) $

3. Differentiate the denominator:
   - $ \frac{d}{d\theta} \sin(5\theta) = 5\cos(5\theta) $

4. Substitute into the limit expression:
   - $ \lim_{\theta \to 0} \frac{2\cos(2\theta)}{5\cos(5\theta)} $

5. Evaluate the limit:
   - Substitute $ \theta = 0 $ into the expression: $ \frac{2\cos(0)}{5\cos(0)} = \frac{2}{5} $

6. The limit of $ \frac{\sin(2\theta)}{\sin(5\theta)} $ as $ \theta $ approaches 0 is $ \frac{2}{5} $.

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

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