e的抬起法运用？

解：lm(x-∞)[(1+1/x)^x]=e^{lm(x-∞)[xln(1+1/x)]} =e^{lm(x-∞)[ln(1+1/x)/(1/x)]} =e^{lm(x-∞)[1/(1+1/x)]} (0/0型，应用罗比达法则) =e^[1/(1+0)] =e.

匿名回答于2023-09-27 12:22:36