Question

L'hopital's rule:
Lim x->0+ [cos((pi/2)-x)]^x

Answer 1

first i would rewrite it as
\[\sin(x)^x\]

Answer 2

then you can either take the log, find the limit, then exponentiate, or write
\[\sin(x)^x=e^{x\sin(x)}\] and take the limit of that one
all the work is teh same either way

Answer 3

That make sense, the answer should come to 0 then. Thanks for you help!

Answer 4

you have to take the limit of \(x\sin(x)\)
that is zero
your limit is \(e^0=1\)