
lim[n→∞] ln(n+1)/lnn =lim[n→∞] [ln(n+1)-lnn+lnn]/lnn =lim[n→∞] [ln(n+1)-lnn]/lnn + 1 =lim[n→∞] ln[(n+1)/n]/lnn + 1 注意:ln[(n+1)/n]/lnn分子极限为0,分母极限是无穷大,因此极限为0 所以原式极限为1. 如果用洛必达法则,需要先将n换成x

lim[n→∞] ln(n+1)/lnn =lim[n→∞] [ln(n+1)-lnn+lnn]/lnn =lim[n→∞] [ln(n+1)-lnn]/lnn + 1 =lim[n→∞] ln[(n+1)/n]/lnn + 1 注意:ln[(n+1)/n]/lnn分子极限为0,分母极限是无穷大,因此极限为0 所以原式极限为1. 如果用洛必达法则,需要先将n换成x
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