Question

求极限，怎样求得结果为e的4次方？

Answer 1

解：∵lim(x->∞)[(4x-1)ln(1+1/x+1/x^2+1/x^3)]
lim(x->∞)[((4-1/x)/(1/x))ln(1+1/x+1/x^2+1/x^3)]
=lim(t->0)[((4-t)/t)ln(1+t+t^2+t^3)] (令t=1/x)
={lim(t->0)(4-t)}*{lim(t->0)[ln(1+t+t^2+t^3)/t]}
=4*{lim(t->0)[ln(1+t+t^2+t^3)/t]}
=4*{lim(t->0)[(1+2t+3t^2)/(1+t+t^2+t^3)]} (0/0型极限，应用罗比达法则)
=4*[(1+0+0)/(1+0+0+0)]
=4
∴原式=lim(x->∞){e^[(4x-1)ln(1+1/x+1/x^2+1/x^3)]} (应用对数性质)
=e^{lim(x->∞)[(4x-1)ln(1+1/x+1/x^2+1/x^3)]} (应用初等函数连续性)
=e^4。