解(1)所求极限是∞/∞未定式,所以lim(x→∞)(3x-5)/(x^2)*sin(1/2x)=lim(x→∞)3/[2xsin(1/2x)+(x^2)cos(1/2x)*(-1/x^2)]=lim(x→∞)3/[2xsin(1/2x)-cos(1/2x)]=-3(2)因为x→0,所以1-cos(x^2)~(x^4)/2,1-cosx~(x^2)/2,lim(x→0)√(1-cos(x^2)/(1-cosx)=lim(x→0)√[(x^4)/2]/(x^2)/2=1.(3)所求极限是0/0未定式,所以lim(x→0)(e-e^cosx)/√(1+x^2)-1=lim(x→0)sinx*e^cosx/(1/2)[(1+x^2)^(-1/2)]*2x=lim(x→0)sinx*e^cosx*(1+x^2)^(1/2)/x=lim(x→0)cosx*e^cosx+sinx*(e^cosx)*(-sinx)*(1+x^2)^(1/2)+sinx*(e^cosx)*(1/2)*[(1+x^2)^(-1/2)]*(2x)=lim(x→0)cos*e^cosx-sin^2(x)*(e^cosx)*(1+x^2)^(1/2)+xsinx*(e^cosx)*(1+x^2)^(-1/2)=lim(x→0)[cosx*(e^cosx)*(1+x^2)^(1/2)-sin^2(x)*(e^cosx)*(1+x^2)+xsinx*e^cosx]/(1+x^2)^(1/2)=e