求极限:lncosax/lncosbxx趋近于零
求极限:lncosax/lncosbx,x趋近于零 -
lim [lncos(ax) / lncos(bx)]x→0 =lim [lncos(ax)]' / [lncos(bx)]'x→0 =lim [-asin(ax)/cos(ax)] / [-bsin(bx)/cos(bx)]x→0 =lim (a/b)[tan(ax)/tan(bx)]x→0 =(a/b)[(ax)/(bx)]=(a/b)(a/b)=a²/b²
采用洛必达法则求导，(lncosax)'=-atanax,(lncosbx)'=-btanbx,所以lim lncosax/lncosbx = (a/b) lim tanax/tanbx,在利用等价无穷小：tanax~ax,tanbx~bx,得到lim lncosax/lncosbx = (a/b) lim tanax/tanbx = a^2/b^2到此结束了？。
lim [lncos(ax) / lncos(bx)]x→0 =lim [lncos(ax)]' / [lncos(bx)]'x→0 =lim [-asin(ax)/cos(ax)] / [-bsin(bx)/cos(bx)]x→0 =lim (a/b)[tan(ax)/tan(bx)]x→0 =(a/b)[(ax)/(bx)]=(a/b)(a/b)=a²/b²
采用洛必达法则求导，(lncosax)'=-atanax,(lncosbx)'=-btanbx,所以lim lncosax/lncosbx = (a/b) lim tanax/tanbx,在利用等价无穷小：tanax~ax,tanbx~bx,得到lim lncosax/lncosbx = (a/b) lim tanax/tanbx = a^2/b^2到此结束了？。
(x->0) lim lncosax/lncosbx =(x->0) lim (cosbx/cosax) * (asinax/bsinbx)=(x->0) lim a²sinax/(ax) / [b²sinbx/(bx)]=a²b²
(lncosax)'=-atanax，lncosbx)'=-btanbx，所以lim lncosax/lncosbx = (a/b) lim tanax/tanbx，在利用等价无穷小：tanax~ax，tanbx~bx，得到 lim lncosax/lncosbx = (a/b) lim tanax/tanbx = a^2/b^2
当x趋近于0时,求lincosax/lincosbx的极限 -
a^2/b^2
ln（cosax－1＋1）ln（cosbx-1+1）(cosax-1)/(cosbx-1)=(-a2x2/2)/(-b2x2/2)=a2/b2
lim(lncosax/lncosbx) -
x->0, ln cos(ax) = ln[ 1+ cos(ax) - 1] 与cos(ax) - 1 是等价无穷小，ln cos(ax) ~ (-1/2)(ax)²ln cos(bx) = ln[ 1+ cos(bx) - 1] 与cos(bx) - 1 是等价无穷小，ln cos(bx) ~ (-1/2)(bx)²lim(x->0) ln cosax／ln cos(bx)是什么。
lim(x→0)[ln(cosax)]/[ln(cosbx)]这是0/0型极限，应用洛必达法则，求导得=lim(x→0)[-(sinax)*a/(cosax)]/[-(sinbx)*b/(cosbx)]=lim(x→0)(tanax)*a/[(tanbx)*b]运用等价无穷小，tanax~ax,tanbx~bx =lim(x→0)ax*a/(bx*b)=a²/b²好了吧！
设a≠0,b≠0,求lim(x→0)[ln(cosax)]/[ln(cosbx)] -
lim(x→0)[ln(cosax)]/[ln(cosbx)]这是0/0型极限，应用洛必达法则，求导得=lim(x→0)[-(sinax)*a/(cosax)]/[-(sinbx)*b/(cosbx)]=lim(x→0)(tanax)*a/[(tanbx)*b]运用等价无穷小，tanax~ax,tanbx~bx =lim(x→0)ax*a/(bx*b)=a²/b²
lim(x→0)[ln(cosax)]/[ln(cosbx)]这是0/0型极限，应用洛必达法则，求导得=lim(x→0)[-(sinax)*a/(cosax)]/[-(sinbx)*b/(cosbx)]=lim(x→0)(tanax)*a/[(tanbx)*b]运用等价无穷小，tanax~ax,tanbx~bx =lim(x→0)ax*a/(bx*b)=a²/b²