洛必达法则求极限lim(lnsinbx/lnsinax)=?x→0lim(lncosbx/lncosax)=?x→0

Question

洛必达法则求极限
lim(lnsinbx/lnsinax)=?
x→0
lim(lncosbx/lncosax)=?
x→0

Answer 1

lim(lnsinbx/lnsinax) 分子分母都是负无穷
=lim b*cosbx*(1/sinbx) / a*cosax*(1/sinax)
=lim btanax / atanbx 分子分母都是0
=lim b[a(1+tanax)] / a[b(1+tanax)]
=ba/ab
=1
lim(lncosbx/lncosax)
=lim b[-sin(bx)]/cos(bx) / a[-sin(ax)]/cos(ax)
=lim btan(bx) / atan(ax)
=lim b {b [1+tan^2 (bx) ] } / a {a [1+tan^2 (ax) ] }
= b^2 *(1+0) / a^2 * (1+0)
=b^2/a^2