f(x) = (1/x)^(sinx) = 1/(x^sinx)
Let us take the limit as x→0+, since rasning negative numbers to real powers is not well defined.
consider limit x→0 g(x)=x^x
lng = x lnx = lnx/(1/x)
limit ln g(x) as x→0 = (1/x)/(-1/x^2) = -x
limit ln(g(x)) = 0
so, limit g(x) = 1
See what you can do with that, since sinx < x
lim x→0 ((1/x)^(sinx))
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