limx^(sinx),x->0+怎么解

Question

limx^(sinx),x->0+怎么解

Answer 1

x^(sinx) = e^[sinxlnx]
lim_{x->0+}[sinxlnx] = lim_{x->0+}[xlnx][sinx/x]
= lim_{x->0+}[xlnx]
= lim_{x->0+}[lnx/(1/x)]
= lim_{x->0+}[(1/x)/(-1/x^2)]
= lim_{x->0+}[-x]
= 0.
所以,
lim_{x->0+}x^(sinx) = e^0 = 1