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KCC's quizzes AQQ255 about a limit at infinite

And here is a new quiz coming!

1. Info: the frequent winners should have received some award and certificate. Please warn me if you think you have been forgotten!

2. Quote of the week: "The elevator to success is out of order. You'll have to use the stairs... one step at a time." - Joe Girard.

3. New quiz of the week: AQQ255 about a limit to infinite

Good luck!



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[edited by: emassa at 3:52 PM (GMT -4) on 21 Mar 2024]
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  • 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.

    •  Analog Employees 
    in reply to TimW

    Fantastic Tim! You get it, congratulations!

  • Thanks  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)

    1. e
    2.  Not really rigorously mathematical, but with a numerical confirmation in Excel, that is "engineeringly" good enough. Here how it goes:

      (1+a)^n      = 1 + na  + n(n-1) a^2  /2!   + n(n-1)(n-2)(n-3)  a^3  ? 3! + ... (binomial expansion)
      (1+1/x)^n   = 1 + n/x  + n(n-1) x^-2  /2! +   ...   (replace a by 1/x )
      (1+1/x)^x    =  1  + 1  +  x(x-1) x^-2  /  2!  +  ...  (replace n  by x )
      (lim  x -> infinity) ((1+1/x)^x)  = 1 + 1 + x^2 x^-2  /2!  +  x^3 x^-3 / 3! +  ...                  
                                                               (  a finite constant is negligeable in front of infinity)
                                                            = e   (Taylor's serie)
  • Correct answer and Excellent devleopement  ; congratulations. An other method could be to use Hospital rule.

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

  • Yes! You get it  ! 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.