E, base of the natural log?

Question

How do we prove that the limit of (1-1/n)^n=1/e as n goes to infinity assuming that we already know that the limit of (1+1/n)^n=e as n goes to infinity?

Answer 1

[n→∞]lim (1-1/n)^n
= [n→∞]lim ((n-1)/n)^n
= [n→∞]lim ((n/(n-1))^(-1))^n
= [n→∞]lim (n/(n-1))^(-n)
= [n→∞]lim ((n/(n-1))^n)^(-1)
= 1/([n→∞]lim (n/(n-1))^n)

Let m=n-1, then m→∞ as n→∞, so this is:

= 1/([m→∞]lim ((m+1)/m)^(m+1))
= 1/([m→∞]lim ((m+1)/m)^(m+1))
= 1/([m→∞]lim (1+1/m)^(m+1))
= 1/(([m→∞]lim (1+1/m)^m)([m→∞]lim (1+1/m)))

We already know that [m→∞]lim (1+1/m)^m = e and [m→∞]lim (1+1/m) can be evaluated directly, so this yields:

= 1/(e*1)
= e^(-1).

So we are done.

Answer 2

e^x = x+1

Such that x is sufficiently small.

Let the value of x be even smaller by replacing with x/y

e^(x/y) = (x/y) + 1

(e^x) = (1+(x/y))^y

So all we do is replace x with anything we want.

x = -1

e^-1 = (1-1/y)^y
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Thank You!