x→-1时求lim(1+1/x)ln(1+x),求右极限
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x→-1时求lim(1+1/x)ln(1+x),求右极限x→-1时求lim(1+1/x)ln(1+x),求右极限x→-1时求lim(1+1/x)ln(1+x),求右极限x→-1时,lim(1+1/x)
x→-1时求lim(1+1/x)ln(1+x),求右极限
x→-1时求lim(1+1/x)ln(1+x),求右极限
x→-1时求lim(1+1/x)ln(1+x),求右极限
x→-1时,
lim(1+1/x)ln(1+x)
=lim[(x+1)ln(1+x)]/x
=-lim[(x+1)ln(1+x)]
=-lim ln(1+x)/[1/(x+1)]
应用洛必达法则
=-lim [1/(1+x)]/[-1/(x+1)^2]
=lim [1/(1+x)]/[1/(x+1)^2]
=lim x+1
=0
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