How do you find #lim cos(3theta)/(pi/2-theta)# as #theta->pi/2# using l'Hospital's Rule?

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Answer 1 · 2017-10-17T16:12:30

Look below

You need to see if the limit is in indeterminate form, so calculate the limit as #theta -> pi/2#

#cos(3(pi/2))/{pi/2-pi/2} = 0/0# which is indeterminate form

now do the derivative of the function

#lim_{theta->pi/2} d/dx[cos(3theta)/{pi/2-theta}]#

#d/dx [cos(3theta)] = 0#

#d/dx [pi/2-theta] = 0#

#0/0# is indeterminate form

so the limit doesn't exist