lim1-cosb(b→∞)为什么不存在? 极限limx→0+e?1xx=( )A.1B.0C.∞D....
tan90\u5ea6\u4e3a\u4ec0\u4e48\u4e0d\u5b58\u5728\uff1f
\u56e0\u4e3atan\u03b8=sin\u03b8/cos\u03b8\uff0c
\u5f53\u03b8=90\u00b0\u65f6\uff0c\u5373tan90\u00b0=sin90\u00b0/cos90\u00b0=1/0\uff0c
\u56e0\u5206\u6bcd\u662f\u4e0d\u80fd\u4e3a0\u7684\uff0c\u6240\u4ee5\u4e0d\u5b58\u5728tan90\u00b0.
limx\u21920+e?1xx\uff1dlimt\u2192\u221etet\uff1dlimt\u2192\u221e1et\uff1d0\u5176\u4e2d\u4ee4t\uff1d1x\u6545\u9009\uff1aB\uff0e
cos函数的值是在-1到1之间而b趋于无穷大的时候
并不能确定cosb的极限值
因为其值并不固定,在-1到1之间波动
所以1-cosb的极限值也不存在
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绛旓細C
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绛旓細/n]]鐨勯暱搴︼紝鏈澶у尯闂撮暱搴ξ=max(xi)鈮^[(1+n)/n]-b^(n/n)=b[b^(1/n)-1]鈫0锛屼笖尉i=b^[(1+2i)/(2n)鈭圼b^(i/n)锛宐^[(1+i)/n]]锛屾弧瓒冲畾绉垎瀹氫箟鐨勬潯浠讹紝鈭村師寮=lim(n鈫掆垶)鈭憑b^[(1+i)/n]-b^(i/n)}sin[b^(2i+1)]=鈭(1,b)sinxdx=cos1-cosb ...
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