Limit

I have already known that limx->infinity (1+1/x)^x=e
But how to find lim x->0 (1+1/x)^x?
Sol
lim(x->0)_(1+1/x)^x
=e^｛ln[lim(x->0)_(1+1/x)^x]｝
=e^｛lim(x->0)_[ln(1+1/x)^x]｝
=e^｛lim(x->0)_[xln(1+1/x)]｝
=e^｛lim(x->0)_[ln(1+1/x)/(1/x)]｝
lim(x->0+)_(1+1/x)^x
=e^｛lim(x->0+)_[ln(1+1/x)/(1/x)]｝
=e^｛lim(y->∞)_[ln(1+y)/y]｝ ∞/∞ type
=e^｛lim(y->∞)_[1/(1+y)]｝ ∞/∞ type
=e^0
=1
lim(x->0-)_(1+1/x)^x
=e^｛lim(x->0-)_[ln(1+1/x)/(1/x)]｝
=e^｛lim(y->-∞)_[ln(1+y)/y]｝ ∞/∞ type
=e^｛lim(y->-∞)_[1/(1+y)]｝ ∞/∞ type
=e^0
=1
So
lim(x->0)_(1+1/x)^x=1

the answer is 1.