∵lim(x->0)[(coshx+cosx-2)/x^4]
=lim(x->0)[(sinhx-sinx)/(4x^3)] (0/0型極限,應(yīng)用羅比達法則)
=lim(x->0)[(coshx-cosx)/(12x^2)] (0/0型極限,應(yīng)用羅比達法則)
=lim(x->0)[(sinhx+sinx)/(24x)] (0/0型極限,應(yīng)用羅比達法則)
=lim(x->0)[(sinhx/x+sinx/x)/24]
=(1+1)/24 (應(yīng)用重要極限lim(x->0)(sin/x)=1,lim(x->0)(sinh/x)=1)
=1/12
∴l(xiāng)im(x->0)[(coshx+cosx-2)/(((sinhx)^2)*((sinx)^2))]
=lim(x->0){[(coshx+cosx-2)/x^4]*[(x/sinhx)^2]*[(x/sinx)^2]}
={lim(x->0)[(coshx+cosx-2)/x^4]}*{[lim(x->0)(x/sinhx)]^2]}*{[lim(x->0)[(x/sinx)]^2}
=(1/12)*(1^2)*(1^2) (應(yīng)用重要極限lim(x->0)(sin/x)=1,lim(x->0)(sinh/x)=1)
=1/12.