当x趋向正无穷时，(2^x+3^x)^(1⼀x)的极限是？

解 因为 lim(2^x+3^x)^(1/x)
>lim(3^x)^1/x
=3
又lim(2^x+3^x)^(1/x)
=lim(2*3^x)1/x
=lim2^1/x*lim(3^x)^1/x
=3
所以当x趋向正无穷时，(2^x+3^x)^(1/x)的极限是3

解法一：原式=lim(x->+∞){3[1+(2/3)^x]^(1/x)}
=3(1+0)^(0)
=3；
解法二：原式=lim(x->+∞){e^[ln(2^x+3^x)/x]}
=e^{lim(x->+∞)[ln(2^x+3^x)/x]}
=e^{lim(x->+∞)[(2^xln2+3^xln3)/(2^x+3^x)]} (∞/∞型极限，应用罗比达法则)
=e^{lim(x->+∞)[((2/3)^xln2+ln3)/((2/3)^x+1)]}
=e^[(0+ln3)/(0+1)]
=e^(ln3)
=3。

y=(2^x + 3^x)^(1/x)
lny = ln(2^x+3^x)/x
应用罗毕达法则，
lny = (2^xln2 + 3^xln3)/(2^x+3^x) = [(2/3)^x ln2 + ln3] / [(2/3)^x +1] = ln3