KCC's quizzes AQQ255 about a limit at infinite

1. This is the constant e (2.71828...)

2. If circular proofs are allowed and we know of the constant e.

Applying limits (1+1/infinite)^(infinite). This means if we rewrite the base as f(x) = (1+1/infinite) and g(x) as the exponent that approaches infinity, we get an indeterminate form as f(x) from the right approaches 1 while the exponent g(x) forces the base to go infinite.

Let h(x) = (1+1/x)^x

Ln h(x) = x * ln(1+1/x)

Reform the right hand side and drop x to the denominator as 1/x.

Ln h(x) = ln(1+1/x)/(1/x)

Apply limits as x approaches infinity and we get 0/0 which is indeterminate.

We can use L'hospital's rule such that:

m(x) = ln(1+1/x)

n(x) = 1/x

m'(x) = -(1/(x^2+x))

n'(x) = -1/x^2

Therefore,

Ln h(x) = (1/(1+1/x))

Apply the limits as x approaches infinity and we get...

Ln h(x) = 1

Since we know ln is log of base e, we can revert it as h(x) = e.

Not a good proof but I'll reply if I can think of a better one.

1. This is the constant e (2.71828...)

2. If circular proofs are allowed and we know of the constant e.

Applying limits (1+1/infinite)^(infinite). This means if we rewrite the base as f(x) = (1+1/infinite) and g(x) as the exponent that approaches infinity, we get an indeterminate form as f(x) from the right approaches 1 while the exponent g(x) forces the base to go infinite.

Let h(x) = (1+1/x)^x

Ln h(x) = x * ln(1+1/x)

Reform the right hand side and drop x to the denominator as 1/x.

Ln h(x) = ln(1+1/x)/(1/x)

Apply limits as x approaches infinity and we get 0/0 which is indeterminate.

We can use L'hospital's rule such that:

m(x) = ln(1+1/x)

n(x) = 1/x

m'(x) = -(1/(x^2+x))

n'(x) = -1/x^2

Therefore,

Ln h(x) = (1/(1+1/x))

Apply the limits as x approaches infinity and we get...

Ln h(x) = 1

Since we know ln is log of base e, we can revert it as h(x) = e.

Not a good proof but I'll reply if I can think of a better one.

Yes! You applied the right approach to prove it in a very formal way! Congratulation NotBeans ! The Hospital method saying Lim[(Fx)/G(x)] = Lim[(F’x)/G’x)] gives quickly the solution.