Wolfram alpha gives a step by step solution here
tps://www.wolframalpha.com/input/?i=lim+as+x+tends+to+0+of+((1%2Bx)%5E(1%2Fx)%2Fe)%5E(1%2Fx)
The link didn't work, I will try again.
https://www.wolframalpha.com/input/?i=lim+as+x+tends+to+0+of+((1%2Bx)%5E(1%2Fx)%2Fe)%5E(1%2Fx)
How using L'Hospital's rule?
limx→0((1+x)1xe)1x
Formula: ln(limx→0f(x))=limx→0lnf(x)
1. We take the logarithm:
ln(limx→0((1+x)1xe)1x)=limx→0ln{((1+x)1xe)1x}=limx→01x⋅ln((1+x)1xe)=limx→01x⋅[ln((1+x)1x)−ln(e)]|ln(e)=1=limx→01x⋅[ln((1+x)1x)−1]=limx→01x⋅[1x⋅ln(1+x)−1]=limx→01x⋅[1x⋅ln(1+x)−xx]=limx→01x⋅[ln(1+x)−xx]=limx→0ln(1+x)−xx2|L'Hospital's rule=limx→0(ln(1+x)−x)′(x2)′=limx→011+x−12x=limx→01−1−x1+x2x=limx→0−x1+x2x=limx→0−x2x(1+x)=limx→0−12(1+x)|x=0=−12(1+0)=−12
2. We revert the logarithm:
ln(limx→0((1+x)1xe)1x)=−12eln(limx→0((1+x)1xe)1x)=e−12limx→0((1+x)1xe)1x=e−12=1e12limx→0((1+x)1xe)1x=1√e
This may be a slightly late response, but yes, that's correct. Thank you.
I did manage to determine the solution in the time that I was waiting for the answer, as the assignment was due the next day, but thank you nonetheless!