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

1 Answer
Noah G
Dec 24, 2016

#1#

Explanation:

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!

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