Math Problem

Find the limit of[(1+1/n)^n－e]*n as n tends to infinite.
Sol
Consider
d(1+x)^(1/x)/dx
Set y=(1+x)^(1/x)
lny=(1/x)ln(1+x)=ln(1+x)/x
d(1+x)^(1/x)/dx
=dy/dx
=(dy/dlny)*(dlny/dx)
=y*(dlny/dx)
=y*｛dln[(1+x)/x]/dx｝
=(1+x)^(1/x)*[1/x*1/(1+x)－ln(1+x)/x^2]
=(1+x)^(1/x)*｛[x－(1+x)ln(1+x)]/[x^2(1+x)]｝
lim(n->∞)_[(1+1/n)^n－e]*n
=lim(x->0+)_[(1+x)^(1/x)－e]/x 0/0 type
=lim(x->0+)_(1+x)^(1/x)*｛[x－(1+x)ln(1+x)]/[x^2(1+x)]｝
=lim(x->0+)_(1+x)^(1/x)*lim(x->0+)_｛[x－(1+x)ln(1+x)]/[x^2(1+x)]｝
=e*lim(x->0+)_｛[x－(1+x)ln(1+x)]/(x^2+x^3)｝
=e*lim(x->0+)_｛[1－ln(1+x)－xln(1+x)]/(x^3+x^2)｝ 0/0 type
=e*lim(x->0+)_｛[－1/(1+x)－x/(1+x)－ln(1+x)]/(3x^2+2)｝
=e*lim(x->0+)_｛[－1－ln(1+x)]/(3x^2+2)｝
=－e/2

Yes, I also know it, but I don't know the details.

This is a good question, I have already bookmarked it.

the answer is -e/2