如题所述
(3)和(5)怎么做埃。(3)我怎么算2啊,答案是0.5埃。,(5)用夹逼定...
答:解:(3)∵lim(x->0)[x/(e^x-1)] =lim(x->0)(1/e^x) (0/0型极限,应用罗比达法则) =1 lim(x->0)[x/ln(1+x)] =lim(x->0)(1+x) (0/0型极限,应用罗比达法则) =1 lim(x->0)[(1-cosx)/x^2] =lim(x->0)[(1/2)(sinx/x)] (0/0型极限,应用罗比达法则)追问
答:解:(3)∵lim(x->0)[x/(e^x-1)] =lim(x->0)(1/e^x) (0/0型极限,应用罗比达法则) =1 lim(x->0)[x/ln(1+x)] =lim(x->0)(1+x) (0/0型极限,应用罗比达法则) =1 lim(x->0)[(1-cosx)/x^2] =lim(x->0)[(1/2)(sinx/x)] (0/0型极限,应用罗比达法则)追问
3题5题我会啊,我是9和11不会啊
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