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

Question

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

Aurora Alvarado · Accepted Answer

#1#

l'Hospital's Rule states that if #lim_(x->c) f(x)/g(x) = 0/0# aka. #f(c) = 0# and #g(c) = 0#, then the limit can be written as #lim_(x->c) (f'(x))/(g'(x))#. If you are still in indeterminate form (#0/0#), then you use l'Hospital's Rule again.

#lim_(theta -> pi/2) (costheta)/(pi/2 - theta) = lim_(theta -> pi/2)(-sintheta)/(-1) = lim_(theta -> pi/2) sin theta = sin(pi/2) = 1#

Hopefully this helps!

Emilia Aguilar · Answer

undefined To find $ \lim_{\theta \to \frac{\pi}{2}} \frac{\cos(\theta)}{\frac{\pi}{2} - \theta} $ using L'Hôpital's Rule, we first take the derivative of the numerator and the derivative of the denominator separately, then evaluate the limit of their ratio.

1. Take the derivative of the numerator:
   $$ \frac{d}{d\theta}(\cos(\theta)) = -\sin(\theta) $$

2. Take the derivative of the denominator:
   $$ \frac{d}{d\theta}\left(\frac{\pi}{2} - \theta\right) = -1 $$

3. Apply L'Hôpital's Rule:
   $$ \lim_{\theta \to \frac{\pi}{2}} \frac{-\sin(\theta)}{-1} $$

4. Evaluate the limit:
   $$ = \lim_{\theta \to \frac{\pi}{2}} \sin(\theta) $$

5. Since $ \sin(\theta) $ approaches 1 as $ \theta $ approaches $ \frac{\pi}{2} $, the limit is 1.

So, $ \lim_{\theta \to \frac{\pi}{2}} \frac{\cos(\theta)}{\frac{\pi}{2} - \theta} = 1 $.

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

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