用洛必达法则求lim(x->0)ln(1+x^2)\(secx-cosx),

用洛必达法则求lim(x->0)ln(1+x^2)\(secx-cosx),
用洛必达法则求lim(x->0)ln(1+x^2)\(secx-cosx),

用洛必达法则求lim(x->0)ln(1+x^2)\(secx-cosx),
这题用等价无穷小代换要简单些
lim(x->0)ln(1+x^2)/(secx-cosx)
=lim(x->0)ln(1+x^2)/(1/cosx-cosx)
=lim(x->0)ln(1+x^2)cosx/(1-(cosx)^2)
=lim(x->0)ln(1+x^2)cosx/((sinx)^2)
等价无穷小代换
=lim(x->0) x^2cosx/x^2
=1
如果非要用洛必达法则,那从倒数第三步
=lim(x->0)ln(1+x^2)cosx/((sinx)^2)
＝lim(x->0)ln(1+x^2)/((sinx)^2)*lim(x->0)cosx
＝lim(x->0) [2x/(1+x^2)]/(2sinxcosx)*1
=lim(x->0) 2x/2sinx*lim(x->0) 1/[(1+x^2)cosx]
=1