作业帮求一道求极限的高数题,lim（x趋近于无穷）[(x+1)/（x-1）]∧x
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求一道求极限的高数题,lim（x趋近于无穷）[(x+1)/（x-1）]∧x
求一道求极限的高数题,lim（x趋近于无穷）[(x+1)/（x-1）]∧x
求一道求极限的高数题,lim（x趋近于无穷）[(x+1)/（x-1）]∧x
lim(x→∞)[(x+1)/(x-1)]^x
=lim(x→∞){[1+2/(x-1)]^[(x-1)/2]}^2*[1+2/(x-1)]
=lim(t→∞)(1+1/t)^2t,其中t=(x-1)/2
=e^2
e^2
为简便省去极限号
分子分母同时除以x,
原式=[(1+1/x)/(1-1/x)]^x=[(1+1/x)]^x/[(1-1/x)]^x=[(1+1/x)]^x乘以[(1-1/x)]^{-x}=e×e=e²
解法一：原式=lim(x->∞){[(1+2/(x-1))^((x-1)/2)]^[2x/(x-1)]}
={lim(x->∞)[(1+2/(x-1))^((x-1)/2)]}^{lim(x->∞)[2x/(x-1)]}
=e^{lim(x->∞)[2x/(x-1)]} (应用重要极限lim(z-...
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解法一：原式=lim(x->∞){[(1+2/(x-1))^((x-1)/2)]^[2x/(x-1)]}
={lim(x->∞)[(1+2/(x-1))^((x-1)/2)]}^{lim(x->∞)[2x/(x-1)]}
=e^{lim(x->∞)[2x/(x-1)]} (应用重要极限lim(z->∞)[(1+1/z)^z]=e)
=e^{lim(x->∞)[2/(1-1/x)]}
=e^[2/(1-0)]
=e²；
解法二：∵lim(x->∞){xln[(x+1)/(x-1)]}=lim(x->∞){[ln(1+1/x)-ln(1-1/x)]]/(1/x)}
=lim(y->0){[ln(1+y)-ln(1-y)]]/y} (令y=1/x)
=lim(y->0)[1/(1+y)+1/(1-y)] (0/0型极限，应用罗比达法则)
=1/(1+0)+1/(1-0)
=2
∴原式=lim(x->∞){e^[xln((x+1)/(x-1))]
=e^{lim(x->∞)[xln((x+1)/(x-1))]}
=e^(2)
=e²。
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