求三題極限

lim(x->0)_｛1/x－1/[x√(x+1)]｝
Sol
Set y=√(x+1)
y^2=x+1
x=y^2－1
lim(x->0)_｛ 1/x－1/[x√(x+1)]｝
=lim(y->1)_｛1/(y^2－1)－1/[(y^2－1)y]｝
=lim(y->1)_｛(y－1)/[(y^2－1)y]｝
=lim(y->1)_｛1/[(y+1)y]｝
=1/2

lim(x->∞)_[√(x^2+3x)－x]
Sol
lim(x->∞)_[√(x^2+3x)－x]
=lim(x->∞)_｛[√(x^2+3x)+x]*[√(x^2+3x)－x]/[√(x^2+3x)+x]｝
=lim(x->∞)_｛[(x^2+3x)－x^2]/[√(x^2+3x)+x]｝
=lim(x->∞)_｛3x/[√(x^2+3x)+x]｝
=lim(x->∞)_｛3/[√(1+3/x)+1/x]｝
=3/2

lim(x->∞)_√x(√(x+6)－√x)
Sol
lim(x->∞)_√x(√(x+6)－√x)
=lim(x->∞)_[√x(√(x+6)－√x)*(√(x+6)+√x)/(√(x+6)+√x)]
=lim(x->∞)_[6√x/(√(x+6)+√x)]
=lim(x->∞)_[6/(√(1+6/x)+√1)]
=3

謝謝提醒，的確應該用"="才對！

William ( 專家 5 級 ):你好。
=lim 1/[(x+1+√(x+1)]
≒1/(1+1)
≒1/2
≒應改為 =
尤其是
1/(1+1)
≒1/2
之≒。

1.lim<x→0>[(1/x-1)/x√(x+1)]≒lim<x→0>[(1/x)/x√(x+1)]=lim<x→0>{1/[x^2*√(x+1)]}≒lim<x→0>[1/(x^2*√1)]=lim<x→0>(1/x^2)=∞........ans
2.lim<x→∞>[√(x^2+3x)-x]≒lim<x→∞>[√(x^2)-x]=lim<x→∞>(x-x)=0........ans
3.lim<x→∞>√x(√(x+6)-√x)=lim<x→∞>{√[x(x+6)]-x}=lim<x→∞>{√(x^2+6x)-x}≒lim<x→∞>[√(x^2)-x]=lim<x→∞>(x-x)=0........ans