x→0+
lim (1+e^(1/x))^x
=lim e^ln (1+e^(1/x))^x
=e^lim ln(1+e^(1/x))^x
考虑
lim ln(1+e^(1/x))^x
=lim xln(1+e^(1/x))
=lim ln(1+e^(1/x)) / (1/x)
换元t=1/x
=lim(t→+∞) ln(1+e^t) / t
该极限为∞/∞型,根据L'Hospital法则
=lim [ln(1+e^t)]' / (t)'
=lim (e^t)/(1+e^t)
=1
故,原极限=e^1=e
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