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当x趋于无穷时,【(x+3)/(x-1)】^x的极限?
1个回答
展开全部
lim(x->无穷大)[(x+3)/(x-1)]^x
对上式取对数
lim(x->无穷大)xln[(x+3)/(x-1)]=xln(x+3)-xln(x-1)=[ln(x+3)-ln(x-1)]/1/x
0/0型用罗必塔法则
=lim(x->无穷大)[1/(x+3)-1/(x-1)]/(-1/x^2)
=lim(x->无穷大)x^2/[1/(x-1)-1/(x+3)]
=lim(x->无穷大)x^2(x+3-x+1)/(x-1)(x+3)
=lim(x->无穷大)4x^2/(x-1)(x+3)
=4
所以lim(x->无穷大)[(x+3)/(x-1)]^x=e^4
对上式取对数
lim(x->无穷大)xln[(x+3)/(x-1)]=xln(x+3)-xln(x-1)=[ln(x+3)-ln(x-1)]/1/x
0/0型用罗必塔法则
=lim(x->无穷大)[1/(x+3)-1/(x-1)]/(-1/x^2)
=lim(x->无穷大)x^2/[1/(x-1)-1/(x+3)]
=lim(x->无穷大)x^2(x+3-x+1)/(x-1)(x+3)
=lim(x->无穷大)4x^2/(x-1)(x+3)
=4
所以lim(x->无穷大)[(x+3)/(x-1)]^x=e^4
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