当x趋近于0时lim e^x+ln(1-x)-1/x-arctanx=?

答案没有错!
原式=lim(x->0){[e^x+1/(x-1)]/[1-1/(1+x²)]} (0/0型极限,应用罗比达法则)
=lim(x->0){(1+x²)*[e^x+1/(x-1)]/x²}
=lim(x->0){(1+x²)*lim(x->0){[e^x+1/(x-1)]/x²}
=(1+0)*lim(x->0){[e^x-1/(x-1)²]/(2x)} (0/0型极限,再次应用罗比达法则)
=lim(x->0){[e^x+2/(x-1)³]/2} (0/0型极限,再次应用罗比达法则)
=[1+2/(0-1)³]/2
=-1/2