求函数极限
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求函数极限
![](http://img.wesiedu.com/upload/d/09/d09fa6d1a25be5a9cc4bcddc17fec4cd.jpg)
![](http://img.wesiedu.com/upload/1/1b/11b12b1679ed1e3072f5997cff2c03e0.jpg)
![](http://img.wesiedu.com/upload/d/09/d09fa6d1a25be5a9cc4bcddc17fec4cd.jpg)
![](http://img.wesiedu.com/upload/1/1b/11b12b1679ed1e3072f5997cff2c03e0.jpg)
原式=(0^2+1)/(0-2*1)=-1/2
三、证明:∵令y=kx时,
lim((x,y)->(0,0))[y/(x+y)]=lim((x,y)->(0,0))[k/(1+k)]=k/(1+k)
这说明取不同的k值极限值就不同
∴lim((x,y)->(0,0))[y/(x+y)]不存在.
原式=lim(t->0)[t/(√(2-e^t)-1)] (令t=xy)
=lim(t->0)[-2√(2-e^t)/e^t] (0/0型极限,应用罗比达法则)
=-2√(2-e^0)/e^0
=-2.
原式=lim((x,y)->(0,0))[(x^2y+9-9)/(x^2y(√(x^2y+9)+3))] (分子分母同乘√(x^2y+9)+3)
=lim((x,y)->(0,0))[1/(√(x^2y+9)+3)]
=1/(√(0^2*0+9)+3)
=1/6.
再问: 3q了,回答的好详细
三、证明:∵令y=kx时,
lim((x,y)->(0,0))[y/(x+y)]=lim((x,y)->(0,0))[k/(1+k)]=k/(1+k)
这说明取不同的k值极限值就不同
∴lim((x,y)->(0,0))[y/(x+y)]不存在.
原式=lim(t->0)[t/(√(2-e^t)-1)] (令t=xy)
=lim(t->0)[-2√(2-e^t)/e^t] (0/0型极限,应用罗比达法则)
=-2√(2-e^0)/e^0
=-2.
原式=lim((x,y)->(0,0))[(x^2y+9-9)/(x^2y(√(x^2y+9)+3))] (分子分母同乘√(x^2y+9)+3)
=lim((x,y)->(0,0))[1/(√(x^2y+9)+3)]
=1/(√(0^2*0+9)+3)
=1/6.
再问: 3q了,回答的好详细