lim^x^2=多少

lim2^nsinx/2^n\u7684\u6781\u9650\u662f\u591a\u5c11


lim [x^2/(x^2-1)]^x
=lim [(x^2-1+1)/(x^2-1)]^x
=lim [1+1/(x^2-1)]^x
=lim {[1+1/(x^2-1)]^(x^2-1) } ^ (x/(x^2-1))
\u5927\u62ec\u53f7\u91cc\u7684\u90e8\u5206\u4e3a\u7b2c\u4e8c\u4e2a\u91cd\u8981\u6781\u9650\uff0c\u7ed3\u679c\u662fe
\u5927\u62ec\u53f7\u5916\u7684\u6307\u6570\u6781\u9650\u662f0
=e^0
=1

\u770b\u6765\u4f60\u5bf9\u7b2c\u4e8c\u4e2a\u91cd\u8981\u6781\u9650\u638c\u63e1\u5f97\u4e0d\u592a\u597d\u5662\uff0c\u8fd9\u9898\u548c\u521a\u624d\u90a3\u9898\u90fd\u662f\u7b2c\u4e8c\u4e2a\u91cd\u8981\u6781\u9650\u7684\u57fa\u672c\u9898\u3002
\u4e3b\u8981\u601d\u8def\u5c31\u662f\u51d1\u6210(1+1/a)^a\u8fd9\u4e2a\u5f62\u5f0f\uff0c\u5176\u4e2da-->\u221e

如果x趋向于0,则答案是0;

如果x趋向于无穷大,则答案是无穷大。

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