Question

求函数极限

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• 系统奖励15（财富值+成长值）+难题奖励20（财富值+成长值）

Answer 1

7.解：∵lim(t->0)[(sint+cost-1)/t]=lim(t->0)[(sint+cost-1)'/t'] (0/0型极限，应用洛必达法则)
==lim(t->0)(cost-sint)=1..........(1)
∴ lim(x->∞){[sin(1/x)+cos(1/x)]^x}=lim(t->0)[(sint+cost)^(1/t)] (令t=1/x)
=lim(t->0){[[1+(sint+cost-1)]^[1/(sint+cost-1))]]^[(sint+cost-1)/t]}
={lim(t->0)[[1+(sint+cost-1)]^[1/(sint+cost-1))]]}^{lim(t->0)[(sint+cost-1)/t]}
=e^{lim(t->0)[(sint+cost-1)/t]} (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e^1 (应用(1))
=e。

Answer 2

令 t=1/x，设原式 = A， 则 t->0
ln(A)
= lim (ln(cos t + sin t))/t
(洛必达)= lim (cos t - sin t)/(cos t + sin t)
= 1
所以 A = e