①设y=e^（1/x）当x＿＿＿时函数为无穷小量，当x＿＿＿时函数为无穷大量。②已知x→0时，ln(1+ax)与sin2x等价，则a=＿＿＿③当x→0时，x^2与sinx...
①设y=e^（1/x）当x＿＿＿时函数为无穷小量，当x＿＿＿时函数为无穷大量。 ②已知x→0时，ln(1+ax)与sin2x等价，则a=＿＿＿ ③当x→0时，x^2与sinx比较是 A.较高级的无穷小量 B.较低级的无穷小量 C.等价无穷小量 D.同阶无穷小量 ④解x→1时lim(x^(1/(1－x))) ⑤解x→0时lim(1/x－1/(e^x－1))
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展开全部.①设y=e^（1/x）当x＿趋于无穷＿＿时函数为无穷小量，当x＿趋于0＿＿时函数为无穷大量。②已知x→0时，ln(1+ax)与sin2x等价即limx→0ln(1+ax)/sin2x=1应用罗比达法则，分子分母同时求导limx→0ln(1+ax)/sin2x=limx→0[a/(1+ax)]/2*（cos2x)=a/2=1,所以a=2③应用罗比达法则，分子分母同时求导limx→0x^2/sinx=lim(x→0)2x/cosx=0所以选A.较高级的无穷小量4、x→1时lim(x^(1/(1－x)))=x→1时limexp(lnx/(1－x))=x→1时limexp(limlnx/(1－x))应用罗比达法则，分子分母同时求导limlnx/(1－x))=lim（-1/x)=-1所以x→1时lim(x^(1/(1－x)))=1/e5、x→0时lim(1/x－1/(e^x－1))=x→0时lim(e^x－1-x)/x*(e^x－1)应用罗比达法则，分子分母同时求导x→0时lim(e^x－1-x)/x*(e^x－1)=x→0时lim(e^x－1)/[(e^x－1+xe^x)再次应用罗比达法则x→0时lim(e^x－1)/[(e^x－1-xe^x)=x→0时lime^x/(2*e^x+xe^x)=1/2',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign),e.getAttribute("jubao"))},getILeft:function(t,e){return t.left+e.offsetWidth/2-e.tip.offsetWidth/2},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#href\}\}/g,e).replace(/\{\{#jubao\}\}/g,n)}},baobiao:{triangularSign:"data-baobiao",tpl:'{{#baobiao_text}}',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign))},getILeft:function(t,e){return t.left-21},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#baobiao_text\}\}/g,e)}}};function l(t){return this.type=t.type
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展开全部洛必达法则=lim1/e^x=1
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展开全部这几题可以用洛必达法则来求,即分子分母同时求导后再求极限： lim(x→0)[e^x-e^(-x)]/(sinx) =lim(x→0)[e^x+e^(-x)]/(cosx) =(e^0+e^0)/cos0 =2 lim(x→0)(x*cot2x) =lim(x→0)(x/tan2x) =lim(x→0)[1*(cos2x)^2/2] =1*(cos0)^2/2 =1/2 lim(x→0){x*[e^(1/x) ]- 1} =lim(x→0)[x/e^(-1/x)-1] =lim(x→0){1/[e^(-1/x)*1/x^2]-1} =lim(x→0)[x^2/e^(-1/x)-1] =-1 lim(x→0)[(1/x)^tanx] =lim(x→0){e^[tanx*ln(1/x)]} =lim(x→0){e^tanx/[1/ln(1/x)]} =lim(x→0){e^{[(cosx)^(-2)]/[x*(-1/x^2)]}} =lim(x→0){e^[-x/(cosx)^2]} =e^(0/1^2) =1 有点长,有问题再问吧……',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign),e.getAttribute("jubao"))},getILeft:function(t,e){return t.left+e.offsetWidth/2-e.tip.offsetWidth/2},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#href\}\}/g,e).replace(/\{\{#jubao\}\}/g,n)}},baobiao:{triangularSign:"data-baobiao",tpl:'{{#baobiao_text}}',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign))},getILeft:function(t,e){return t.left-21},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#baobiao_text\}\}/g,e)}}};function l(t){return this.type=t.type
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