1.lim(x→0) x.Sin1/x
2.lim(x→0)(1+x/2)^-1/x
3.lim(x→∞)(x+5)/(x-5) ^x
4.lim(x→0)xcotx
5.lim(x→0+)(cosx-1)/(x^1.5)
第1个回答 2008-10-27
1,lim(x→0) x*sin1/x=(sin1/x)*lim(x→0)x=0
2,ln(1+x/2)^-1/x =(-1/x)ln(1+x/2),当x趋向于0时,x和ln(1+x/2)是等价的,即lim(x→0)ln(1+x/2)^-1/x=1,所以lim(x→0)(1+x/2)^-1/x =0
3,lim(x→∞)(x+5)/(x-5) ^x=lim(x→∞)(x+5)/(x-5)*lim(x→∞)1/(x-5) ^(x-1)=1*0=0
4,lim(x→0)xcotx=lim(x→0)xcosx/sinx=lim(x→0)x/sinx*lim(x→0)cosx=1*1=1
5,当x→0+时,分子分母都趋向于0,用洛比达法则,上下求导,分子cosx-1比分母x^1.5高阶,因此lim(x→0+)(cosx-1)/(x^1.5)=0
2,ln(1+x/2)^-1/x =(-1/x)ln(1+x/2),当x趋向于0时,x和ln(1+x/2)是等价的,即lim(x→0)ln(1+x/2)^-1/x=1,所以lim(x→0)(1+x/2)^-1/x =0
3,lim(x→∞)(x+5)/(x-5) ^x=lim(x→∞)(x+5)/(x-5)*lim(x→∞)1/(x-5) ^(x-1)=1*0=0
4,lim(x→0)xcotx=lim(x→0)xcosx/sinx=lim(x→0)x/sinx*lim(x→0)cosx=1*1=1
5,当x→0+时,分子分母都趋向于0,用洛比达法则,上下求导,分子cosx-1比分母x^1.5高阶,因此lim(x→0+)(cosx-1)/(x^1.5)=0