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lim x→∞ [(x+3)\/(x-1)]^(x+1)过程
你好！答案是e^4，如图改写函数的形式就可以利用特殊极限。经济数学团队帮你解答，请及时采纳。谢谢！
求极限limx→∞[(3x+4)\/(3x-1)]^(x+1)=?
[(3x+4)\/(3x-1)]^(x+1)=[1+ 5\/(3x-1)]^(x+1)= [1+ 5\/(3x-1)]^(x-1\/3) * [1+ 5\/(3x-1)]^(4\/3)= [1+ 5\/(3x-1)]^(4\/3) * {[1+ 5\/(3x-1)]^[(3x-1)\/5)]}^(5\/3) =e^(5\/3)
求x趋于无穷((x+3)÷(x+1))^(x+1)的极限
lim((x+3)÷(x+1))^(x+1)=lim{(x+1+2)\/(x+1)}^(x+1)=lim[1+2\/(x+1)]^(x+1)=e^{lim[2\/(x+1)](x+1)}=e^lim[2(x+1)\/(x+1)]=e^2
极限lim(x趋近于∞)[(x+3)\/(x-1)]^x等于多少?
lim（x→∞）[（x+3）\/（x-1）]^x =lim（x→∞）[1+4\/（x-1）]^x =lim（x→∞）[1+4\/（x-1）]^(x-1)*lim（x→∞）[1+4\/（x-1）]=lim（x→∞）{[1+4\/（x-1）]^[(x-1)\/4]}^4 =e^4
当x趋于无穷时,【(x+3)\/(x-1)】^x的极限?
\/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 ...
求极限 lim(x→∞)(x+3\/x-1)x次幂
简单计算一下即可，答案如图所示
lim x→∞ [(x-1)\/(x+3)]^(x+1)
lim[x->oo] (1+x³)^(1\/4) \/ (1+x) = lim [(1+x³) \/ (1+x)^4]^(1\/4)，将分母移进根号内 = lim {(1\/x^4+1\/x)\/[(1+x)\/x]^4}^(1\/4)，上下分别除以x^4 = lim [(1\/x^4+1\/x)\/(1+1\/x)^4]^(1\/4) = [(0+0)\/(1+0)]^(1\/4) = 0...
limx→∞(x+3\/x+1)^(x+2)=
详细过程如图所示，希望能帮到你，解决你所需要的问题
...请问这个式子的极限是多少 lim [(x+3)÷(x-1)]^(x+2) x→∞...
这么写看懂吗?
lim(x趋于∞)[(x+2)\/(x+1)]^(3x+1)=( )
lim(x趋于∞)(1+1\/x)^x=e,lim(x+1趋于∞)[1+1\/(x+1)]^(x+1)=e,lim(x+1趋于∞)[1+1\/(x+1)]=1 lim(x趋于∞)[1+1\/(x+1)]^[3(x+1)-2]=lim(x趋于∞){[1+1\/(x+1)]^[3(x+1)]}{[1+1\/(x+1)]^(-2)} =lim(x趋于∞){[1+1\/(x+1)]^[3(x+1)]...
