How do you find the limit of #(1 - 1/x)^x# as x approaches infinity?

#lim_(xrarroo)(1-1/x)^x# has the form #1^oo# che è una forma indeterminata.

We will use logarithms and the exponential function.

Adesso,
#(1-1/x)^x = e^(ln(1-1/x)^x)#

So we will investigate the limit of the exponent.

#lim_(xrarroo)(ln(1-1/x)^x)#

It will be convenient to note that: #1-1/x = (x-1)/x#

#ln(1-1/x)^x = ln ((x-1)/x)^x = xln((x-1)/x)#

(Using a property of logarithms to bring the exponent down)

Adesso come #xrarroo# we get the form # oo * ln1 = oo*0# So we'll put the reciprocal of one of these in the denominator so we can use l'Hopital's Rule.

#xln((x-1)/x) = (ln((x-1)/x))/(1/x)# Ora, come #xrarroo# we get the form #0/0# Apply l'Hopitals's Rule:

The more tedious derivative is:
#d/dx(ln((x-1)/x)) = 1/((x-1)/x)*d/dx((x-1)/x)#

# = x/(x-1) * 1/x^2 = 1/(x(x-1))#

So we get from #(ln((x-1)/x))/(1/x)#

a #(1/(x(x-1)))/(-1/x^2) = - x/(x-1)#

Now we can probably find this limit without l'Hopital:

#lim_(xrarroo)(- x/(x-1)) = -1#

Sommario:

#(1-1/x)^x = e^(ln(1-1/x)^x)#

E come #xrarroo# the exponent goes to #-1#

Perciò:

#lim_(xrarroo)(1-1/x)^x = lim_(xrarroo)e^(ln(1-1/x)^x) = e^-1 = 1/e#