大一高数求解.lim {(1+n)[In(1+n)-In(n)]}.n趋向无穷.
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大一高数求解.lim{(1+n)[In(1+n)-In(n)]}.n趋向无穷.大一高数求解.lim{(1+n)[In(1+n)-In(n)]}.n趋向无穷.大一高数求解.lim{(1+n)[In(1+
大一高数求解.lim {(1+n)[In(1+n)-In(n)]}.n趋向无穷.
大一高数求解.lim {(1+n)[In(1+n)-In(n)]}.n趋向无穷.
大一高数求解.lim {(1+n)[In(1+n)-In(n)]}.n趋向无穷.
lim(n→∞) {(1+n)[In(1+n)-In(n)]
=lim(n→∞) (1+n)In[(1+n)/n]
=lim(n→∞) (1+n)In[1+1/n)
=lim(n→∞) (1+n)*1/n
=1
ln(1+n)-lnn=ln(1+1/n)
然后令t->0+ t=1/n
化成0/0型，就可以用洛必达法则做了