Question

求极限 lim (coshx+cosx-2)/{[(sinhx)^2][(sinx)^2)]},x趋近于0

Answer 1

∵lim(x->0)[(coshx+cosx-2)/x^4]

=lim(x->0)[(sinhx-sinx)/(4x^3)] (0/0型极限,应用罗比达法则)

=lim(x->0)[(coshx-cosx)/(12x^2)] (0/0型极限,应用罗比达法则)

=lim(x->0)[(sinhx+sinx)/(24x)] (0/0型极限,应用罗比达法则)

=lim(x->0)[(sinhx/x+sinx/x)/24]

=(1+1)/24 (应用重要极限lim(x->0)(sin/x)=1,lim(x->0)(sinh/x)=1)

=1/12

∴lim(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) (应用重要极限lim(x->0)(sin/x)=1,lim(x->0)(sinh/x)=1)

=1/12.