Question

求极限[（2^1/x+3^1/x+4^1/x）/3]^x

Answer 1

令t=1/x，则原式=lim(t->0) [(2^t+3^t+4^t)/3]^(1/t)
（自然对数代换）=lim(t->0) e^{(1/t)*ln[(2^t+3^t+4^t)/3]}
（洛必达法则）=lim(t->0) e^[3/(2^t+3^t+4^t)*(ln2*2^t+ln3*3^t+ln4*4^t)/3]
=lim(t->0) e^[(ln2*2^t+ln3*3^t+ln4*4^t)/(2^t+3^t+4^t)]
=e^[(ln2+ln3+ln4)/3]
=e^(ln24/3)
=24^(1/3)