求lim(n^p+n^q)^1/n
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求lim(n^p+n^q)^1/n求lim(n^p+n^q)^1/n求lim(n^p+n^q)^1/nlim(n->∞)(n^p+n^q)^(1/n)=e^limln(n^p+n^q)/n=e^lim
求lim(n^p+n^q)^1/n
求lim(n^p+n^q)^1/n
求lim(n^p+n^q)^1/n
lim(n->∞) (n^p+n^q)^(1/n)
=e^lim ln(n^p+n^q)/n
=e^lim 1/[pn^(p-1)+qn^(q-1)],洛必达法则
=e^(1/∞)
=e^0
=1
求lim(n^p+n^q)^1/n
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