Question

求(1+2^x+3^x)^1/x当x趋于正无穷时的极限

Answer 1

解：∵lim(x->+∞)[ln(1+2^x+3^x)/x]
=lim(x->+∞)[(ln(1+2^x+3^x))'/(x)'] (∞/∞型极限，应用罗比达法则)
=lim(x->+∞)[(2^x*ln2+3^x*ln3)/(1+2^x+3^x)]
=lim(x->+∞)[((2/3)^x*ln2+ln3)/((1/3)^x+(2/3)^x+1)]
=(0*ln2+ln3)/(0+0+1)
=ln3
∴lim(x->+∞)[(1+2^x+3^x)^(1/x)]
=lim(x->+∞){(e^[ln(1+2^x+3^x)/x]}
=e^{lim(x->+∞)[ln(1+2^x+3^x)/x]}
=e^3。

追问

最后一步写错啦！

Answer 2

3^x < 1+2^x+3^x < 3 * 3^x
3 < (1+2^x+3^x) ^ (1/x) < 3^(1/x) * 3
lim(x->+∞) 3^(1/x) = 1
由迫敛准则（夹挤准则）得：
lim(x->+∞) (1+2^x+3^x) ^ (1/x) = 3

Answer 3

先取对数，再用洛比达法则，lim [ln(1+2^x+3^x)]/x=lim(2^xln2+3^xln3)/(1+2^x+3^x)=ln3
所以lim(1+2^x+3^x)^1/x=e^(ln3)=3 (x趋于正无穷）