求解第5题和第6题
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（5）原式=lim(x->0) {[1+(a^x+b^x+c^x-3)/3]^[3/(a^x+b^x+c^x-3)]}^[(a^x+b^x+c^x-3)/3x]
=e^lim(x->0) [(a^x+b^x+c^x-3)/3x]
=e^lim(x->0) (lna*a^x+lnb*b^x+lnc*c^x)/3
=e^[ln(abc)/3]
=(abc)^(1/3)
（6）原式=lim(x->π/2) {(1+sinx-1)^[1/(sinx-1)]}^[tanx(sinx-1)]
=e^lim(x->π/2) [tanx(sinx-1)]
=e^lim(x->π/2) (sin^2x-sinx)/cosx
=e^lim(x->π/2) (2sinxcosx-cosx)/(-sinx)
=e^0
=1
=e^lim(x->0) [(a^x+b^x+c^x-3)/3x]
=e^lim(x->0) (lna*a^x+lnb*b^x+lnc*c^x)/3
=e^[ln(abc)/3]
=(abc)^(1/3)
（6）原式=lim(x->π/2) {(1+sinx-1)^[1/(sinx-1)]}^[tanx(sinx-1)]
=e^lim(x->π/2) [tanx(sinx-1)]
=e^lim(x->π/2) (sin^2x-sinx)/cosx
=e^lim(x->π/2) (2sinxcosx-cosx)/(-sinx)
=e^0
=1
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