Question

lim x goes to (pi/2), [ln(sinx) + x - (pi/2)] / [cosx]

Answer 1

\[\lim_{x\rightarrow \frac{\pi}{2}} \frac{\ln(\sin(x))+x-\frac{\pi}{2}}{\cos(\frac{\pi}{2})}\] this one?

Answer 2

or rather i meant
\[\lim_{x\rightarrow \frac{\pi}{2}} \frac{\ln(\sin(x))+x-\frac{\pi}{2}}{\cos(x)}\]

Answer 3

yes thats the one, i got -2 as an answer

is that right? i used l'hospitals rule

Answer 4

get 0/0 straight up l'hopital problem. take the derivative top and bottom get
\[\frac{\frac{\cos(x)}{\sin(x)}+1}{-\sin(x)}\]

Answer 5

replace x by
\[\frac{\pi}{2}\] get -1

Answer 6

not sure where the -2 came from unless you forgot the chain rule when taking the derivative of
\[\ln(\sin(x))\]