lim(x->+∞)[3^x+2^x]/[(3^x+1)+(2^x+1)]

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lim(x->+∞)[3^x+2^x]/[(3^x+1)+(2^x+1)]
lim(x->+∞)[3^x+2^x]/[(3^x+1)+(2^x+1)]

lim(x->+∞)[3^x+2^x]/[(3^x+1)+(2^x+1)]
分子分母同除以3^x,得:
原式=[1+(2/3)^x]/[3+2×(2/3)^x]
===>>> [1]/[3]=1/3

只看最高次项3^x的系数比
结果是1/3

lim(x->+∞)[3^x+2^x]/[(3^x+1)+(2^x+1)]=lim(x->+∞)1-2/[(3^x+1)+(2^x+1)]=1

lim(x->+∞)[3^x+2^x]/[(3^x+1)+(2^x+1)]=lim(x->+∞)[-2]/[(3^x+1)+(2^x+1)]=0

上下同时除以3^x,则(2/3)^x趋于0。所以答案为1/3。懂了吗?

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