n是自然数,当n趋于无穷大时,求[n·tan(1/n)]^(n^2)的极限james_miao的详细解答,但正确答案确实是e^(1/3)...
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![n是自然数,当n趋于无穷大时,求[n·tan(1/n)]^(n^2)的极限james_miao的详细解答,但正确答案确实是e^(1/3)...](/uploads/image/z/7833015-63-5.jpg?t=n%E6%98%AF%E8%87%AA%E7%84%B6%E6%95%B0%2C%E5%BD%93n%E8%B6%8B%E4%BA%8E%E6%97%A0%E7%A9%B7%E5%A4%A7%E6%97%B6%2C%E6%B1%82%5Bn%C2%B7tan%281%2Fn%29%5D%5E%28n%5E2%29%E7%9A%84%E6%9E%81%E9%99%90james_miao%E7%9A%84%E8%AF%A6%E7%BB%86%E8%A7%A3%E7%AD%94%EF%BC%8C%E4%BD%86%E6%AD%A3%E7%A1%AE%E7%AD%94%E6%A1%88%E7%A1%AE%E5%AE%9E%E6%98%AFe%5E%281%2F3%29...)
n是自然数,当n趋于无穷大时,求[n·tan(1/n)]^(n^2)的极限james_miao的详细解答,但正确答案确实是e^(1/3)...
n是自然数,当n趋于无穷大时,求[n·tan(1/n)]^(n^2)的极限
james_miao的详细解答,但正确答案确实是e^(1/3)...
n是自然数,当n趋于无穷大时,求[n·tan(1/n)]^(n^2)的极限james_miao的详细解答,但正确答案确实是e^(1/3)...
为了计算方便,令x=1/n,则n趋于无穷时,x趋于0,原式变形为求(tanx/x)^(1/x^2)的极限
而原式=lim e^[(1/x^2)*ln(tanx/x)]
这样,我们只需要求出x趋于0时,指数部分[(1/x^2)*ln(tanx/x)]的极限即可
观察ln(tanx/x)/(x^2),属0/0型,运用洛比达法则
lim ln(tanx/x)/(x^2)=lim [x*(secx)^2-tanx]/[2tanx*(x^2)],
0/0型,洛比达
lim [x*(secx)^2-tanx]/[2tanx*(x^2)]=lim [x*tanx*(secx)^2]/[2xtanx+x^2*(secx)^2]
分子分母同时除以x*(secx)^2,得
原式=lim [tanx]/[sin2x+x]
=lim [(secx)^2]/[2cos2x+1](洛比达)
=1/3
即x趋于0时,[(1/x^2)*ln(tanx/x)]的极限为1/3,lim e^[(1/x^2)*ln(tanx/x)]=e^(1/3)
所以当n趋于无穷大时,[n·tan(1/n)]^(n^2)的极限为e^(1/3)
……不知道有没手误打错了,
用maple算得
> limit((n*tan(1/n))^(n^2), n = infinity);
=e^(1/3)
n=∞
1/n = 1/∞ = 0
tan(0) = 0
n*tan(0) = n*0 = 0
0^任何数 = 0
[n·tan(1/n)]^(n^2) = (n*0)^(n^2) = 0^(n^2) = 0
[n·tan(1/n)]^(n^2)的极限 = 0
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复制别人的答案晚上会尿床的,各位小孩儿们好自为之