大一高数，用图中提示证明当n→∝时lim n√n＝1
\u5927\u4e00\u9ad8\u6570\u8bc1\u660e\u9898\uff1a\u82e5an>0,\u4e14lim(n\u2192\u221e)a(n+1)/a(n)=a\uff0c\u5219lim\uff08an^(1/n)\uff09=a\u5229\u7528stolz\u5b9a\u7406\uff0c\u662f\u6700\u7b80\u5355\u7684\u505a\u6cd5
\u7ed3\u8bba\u662f\u660e\u663e\u7684~~~
\u5982\u679c\u4e0d\u7528stolz\u5b9a\u7406\uff0c\u505a\u6cd5\u5176\u5b9e\u4e5f\u4e0d\u96be~~
lim(n\u2192\u221e)a(n+1)/a(n)=a
\u6839\u636e\u5b9a\u4e49\uff1a
\u5bf9\u4efb\u610f\u03b5>0\uff0c\u5b58\u5728N>0\uff0c\u5f53N>N\uff0c\u5c31\u6709|a(n+1)/a(n)-a|<\u03b5
\u5373\u6709\uff1a(a-\u03b5)<a(n+1)/an<(a+\u03b5)
(a-\u03b5)<an/a(n-1)<(a+\u03b5)
(a-\u03b5)<a(n-1)/a(n-1)<(a+\u03b5)
\u2026\u2026
(a-\u03b5)<a(N+2)/a(N+1)<(a+\u03b5)
\u56e0\u4e3a\u5404\u9879\u540c\u4e3a\u6b63\u6570\uff0c\u76f8\u4e58\uff1a
(a-\u03b5)^(n-N-1)<an/a(N+1)<(a+\u03b5)^(n-N-1)
\u540c\u65f6\u5f00n\u6b21\u65b9\uff1a
(a-\u03b5)^[(n-N-1)/n]<(an/a(N+1))^(1/n)<(a+\u03b5)^[(n-N-1)/n]
\u5bf9\u4e0a\u5f0f\u540c\u65f6\u53d6\u6781\u9650\uff1a
lim(n\u2192\u221e) (a-\u03b5)^[(n-N-1)/n]\u2264lim(n\u2192\u221e) (an/a(N+1))^(1/n)\u2264lim(n\u2192\u221e) (a+\u03b5)^[(n-N-1)/n]
(a-\u03b5)\u2264lim(n\u2192\u221e) an^(1/n)\u2264(a+\u03b5)
\u5373\uff1a|lim(n\u2192\u221e) an^(1/n) - a| \u2264 \u03b5
\u4e5f\u5c31\u662f\u8bf4\uff0c\u5bf9\u4efb\u610f\u03b5>0\uff0c\u90fd\u6709|lim(n\u2192\u221e) an^(1/n) - a| \u2264 \u03b5
\u6545\uff0clim(n\u2192\u221e) an^(1/n) = a
\u6709\u4e0d\u61c2\u6b22\u8fce\u8ffd\u95ee
lim(1+1/n+1/n^2)^n=lim(1+\uff08n+1\uff09/n^2)^n
lim(1+\uff08n+1\uff09/n^2)^\uff08n^2/\uff08n+1\uff09\uff09=e
=lim(1+\uff08n+1\uff09/n^2)^\uff08\uff08n^2-1+1\uff09/\uff08n+1\uff09\uff09
\u56e0\u4e3alim(1+\uff08n+1\uff09/n^2)^\uff081/\uff08n+1\uff09\uff09=1
\u6240\u4ee5\u539f\u5f0f=lim(1+\uff08n+1\uff09/n^2)^\uff08n-1\uff09\u4e58\u4ee5lim(1+\uff08n+1\uff09/n^2)^\uff081/\uff08n+1\uff09\uff09
\u6240\u4ee5lim(1+\uff08n+1\uff09/n^2)^\uff08n-1\uff09=e
\u6240\u4ee5lim(1+1/n+1/n^2)^\uff08n-1\uff09=e
\u6240\u4ee5lim(1+1/n+1/n^2)^n=e
lim(1+（n+1）/n^2)^（n^2/（n+1））=e
=lim(1+（n+1）/n^2)^（（n^2-1+1）/（n+1））
因为lim(1+（n+1）/n^2)^（1/（n+1））=1
所以原式=lim(1+（n+1）/n^2)^（n-1）乘以lim(1+（n+1）/n^2)^（1/（n+1））
所以lim(1+（n+1）/n^2)^（n-1）=e
所以lim(1+1/n+1/n^2)^（n-1）=e
所以lim(1+1/n+1/n^2)^n=e
河边两只鹅，
绛旓細鎵浠im(1+1/n+1/n^2)^n=e
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