lim(n趋向于无穷) (cosx/n+sinx/n)^n 的值为多少?
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∵lim(t->0)[(sint+cost-1)/t]=lim(t->0)(cost-sint) (0/0型极限,应用罗比达法则)
=1
∴ lim(n->∞){[cos(x/n)+sin(x/n)]^n}
=lim(t->0)[(cost+sint)^(x/t)] (令t=x/n)
=lim(t->0){[(1+(sint+cost-1)]^[(1/(sint+cost-1))*(x(sint+cost-1)/t)]}
=e^{x*lim(t->0)[(sint+cost-1)/t]} (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e^(x*1)
=e^x.
=1
∴ lim(n->∞){[cos(x/n)+sin(x/n)]^n}
=lim(t->0)[(cost+sint)^(x/t)] (令t=x/n)
=lim(t->0){[(1+(sint+cost-1)]^[(1/(sint+cost-1))*(x(sint+cost-1)/t)]}
=e^{x*lim(t->0)[(sint+cost-1)/t]} (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e^(x*1)
=e^x.
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