Euler's Number

By definition, Euler's number (e) is the limit as n increases without bound of
(1+

)ⁿ, or the limit as v approaches 0 from the positive direction of (1+v)^1
v, approximately 2.71828182846.

What is the limit, as n increases without bound, of (1+

)ⁿ? Derive your solution without the tools of Calculus.

View the solution

Euler's Number Solution

x = lim n -> ∞ (1+

)ⁿ = lim n -> ∞ (

n+2

)ⁿ

e = lim n -> ∞ (1+

)ⁿ = lim n -> ∞ (

n+1

)ⁿ

x
e
= lim n -> ∞ (

n+2

n+1

)ⁿ.

Let u = n+1.

lim n -> ∞ (

n+2

n+1

)ⁿ = lim u -> ∞ (

u+1

)^u-1

But as u increases without bound, u-1 is equivalent to u -- so,
lim u -> ∞ (

u+1

)^u-1 = lim u-> ∞ (

u+1

)^u = e

Therefore, x
e
= e, so

x = e²

Thanks, Soroban

Easy as 2, 3, 4

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