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
lim (1+x)1/x - e / x = ?
x\rightarrow0
here d numerator is whole divided by denominator(for any confusion)
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
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