limiting limits

i'm having problem with L'hospital, dude as answer is given -e/2

abhi ................. here put the expansion of [1+x][/x ]

u will get answer -e/2

yup d ans is -e/2

1st u apply L-hospital. and then by expanding ln(1+x) and cancelling x² from numerator and denominator u get -e/2

direct expansion pf (1+x)^1/x is a better method i guess

can u explain that in detail

lim x-->0 (1+x) raise to 1/x

nw wen x-->0 ....then 1+x-->1

and wen x-->0 then 1/x -->infi

so itz of type 1 raise to infi
so lim x-->0 (1+x)^(!/x)

e^(limx-->0(1+x-1) *(1/x)

=e
solve d second limt n u get d answer

yes i 2 got -e/2