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

#### Risposta:

The limit is 1/e

#### Spiegazione:

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

We will use logarithms and the exponential function.

(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