高数求极限求详解
2个回答
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lim(x->1)[x/(x-1)-1/lnx]
=lim(x->1)[(xlnx-(x-1))/(x-1)lnx]
=lim(x->1)[(xlnx-x+1)/(xlnx-lnx)]
=lim(x->1)[((lnx+1)-1)/(lnx+1)-1/x]←洛必达法则
=lim(x->1){lnx/[(xlnx+x-1)/x]}
=lim(x->1)[xlnx/(xlnx+x-1)]
=lim(x->1)[(lnx+1)/(lnx+2)]←洛必达法则
=(0+1)/(0+2)
=1/2
=lim(x->1)[(xlnx-(x-1))/(x-1)lnx]
=lim(x->1)[(xlnx-x+1)/(xlnx-lnx)]
=lim(x->1)[((lnx+1)-1)/(lnx+1)-1/x]←洛必达法则
=lim(x->1){lnx/[(xlnx+x-1)/x]}
=lim(x->1)[xlnx/(xlnx+x-1)]
=lim(x->1)[(lnx+1)/(lnx+2)]←洛必达法则
=(0+1)/(0+2)
=1/2
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