Question

lim cosx/(1-sinx)
x->(pi/2)^+

Answer 1

looks like \(\frac{0}{0}\) l'hopital should cure that in one step

Answer 2

take the derivative top and bottom get
\[\frac{-\sin(x)}{-\cos(x)}=\frac{\sin(x)}{\cos(x)}\]

Answer 3

now it is of the form \(\frac{1}{0}\) so there is no limit

Answer 4

yes, 0/0. then using L'H i get -sinx/-cosx which equals tanx. then if i plug in pi/2 for x, i get an error on my calc. the answer in the back says neg infinity

Answer 5

the back of the book is full of it. there is no limit. but since you are going to \(\frac{\pi}{2}\) from above, then the curve is going to minus infinity for sure

Answer 6

hmm