Use l'Hospital's Rule to show that lim x-> infinity (1+1/x)^x =e
lim (1 + 1/x)^x
log lim = lim x log(1 + 1/x)
= lim log(1 + 1/x) / (1/x)
= lim (-1/(x^2+x)) / (-1/x^2)
= lim x^2/(x^2+x)
= 1
so, log lim = 1
lim = e^1 = e
log lim = lim x log(1 + 1/x)
= lim log(1 + 1/x) / (1/x)
= lim (-1/(x^2+x)) / (-1/x^2)
= lim x^2/(x^2+x)
= 1
so, log lim = 1
lim = e^1 = e