KCC's quizzes AQQ255 about a limit at infinite

1)  It is the math constant 'e'

2)  y=\lim_{x\to\text{infinity}}(1+\frac{1}{x})^x

\ln(y)=\lim_{x\to\text{infinity}}x\ln(1+\frac{1}{x})

rearrange to:

\ln(y)=\lim_{x\to\text{infinity}}\ln\frac{(1+\frac{1}{x})}{\frac{1}{x}}

This limit leads to 0/0, so we apply l'Hopital's rule and take derivative

\frac{-\frac{1}{x^2}}{\frac{(1+\frac{1}{x})}{-\frac{1}{x^2}}}

Take the limit again, which gives:

\ln(y)=\lim_{x\to\imaginaryI nf}(1+\frac{1}{x})

\ln(y)=1

e^{\ln(y)}=e^1

y=e

As x increases toward infinity, the 1/x term approaches zero. The term in parenthesis approaches 1, e.g. (1+1/x) -> (1)

1 raised to any power is 1, so regardless of the power of x, 1 raised to the x power is 1.

So when x is large the limit of this equation is approximately 1.

Fantastic Tim! You get it, congratulations!

Thanks rsprang for your prompt reply. But I think you have been a little bit too fast this time..It is correct to say 1/x approaches 0 and 1+1/x will tend to 1. BUT, in parallel, the exponent will put the whole value to infinite. The question will be then who goes faster? The exponent or the 1/x ?

Yes, that did seem a little too simple, I wondered if I was missing something......

Thanks for the comment and suggestions.

(oops! I send through email rather that here.  I repost it here just for a proper way to do it)

Correct answer and Excellent devleopement vanderghast ; congratulations. An other method could be to use Hospital rule.

From above the limit of ln [f(x)] equals 1.

Yes! You get it retiredEE ! Congratulations!

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.