当x→1时,求x/x-1-1/lnx的极限?求 lim ( x/x-1 - 1/lnx )的极限?x→1

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当x→1时,求x/x-1-1/lnx的极限?求lim(x/x-1-1/lnx)的极限?x→1当x→1时,求x/x-1-1/lnx的极限?求lim(x/x-1-1/lnx)的极限?x→1当x→1时,求x

当x→1时,求x/x-1-1/lnx的极限?求 lim ( x/x-1 - 1/lnx )的极限?x→1
当x→1时,求x/x-1-1/lnx的极限?
求 lim ( x/x-1 - 1/lnx )的极限?
x→1

当x→1时,求x/x-1-1/lnx的极限?求 lim ( x/x-1 - 1/lnx )的极限?x→1
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

通分后是0/0型用洛比达法则,分子分母求两次导后得1/2

x/(x-1)-1/lnx=(xlnx-x+1)/(x-1)lnx=(xlnx-x+1)/(x-1)ln(1+x-1)=(xlnx-x+1)(x-1)/(x-1)=(xlnx-x+1)=0注:x-1与ln(1+x-1)为等价无穷小,省略了lim