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Give the Limit

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data

    give the limit if it exist of (1-1/n)^n n=1 to infinity

    the teacher wrote:
    lim n goes to infinity (1+x/n)^n = e^x

    so therefore
    lim n goes to infinity (1-1/n)^n = e^-1

    can someone explain this to me.
     
  2. jcsd
  3. Mar 31, 2009 #2
    All they've done is given x a value. And since the two expressions are equivalent, when x is a specific value in one expression, to remain equivalent x has the same value in the other.

    So all your teacher did was substitute -1 for x?
     
  4. Mar 31, 2009 #3
    how i'm i suppose to know that

    (1+x/n)^n = e^x ??
     
  5. Mar 31, 2009 #4
    Ah, yes I was wondering if that was your question.

    Well the proof exists online... it's not too terribly difficult.

    To prove that (where x is some constant) [tex]\mathop{\lim }\limits_{n \to \infty } (1+\frac{x}{n})^{n} = e^x[/tex]

    We first set the limit equal to an arbitrary variable (y for instance).

    [tex]y =\mathop {\lim }\limits_{n \to \infty } (1+\frac{x}{n})^{n}[/tex]

    Next we'll utilize the rules of the natural logarithm to bring down the power from the quantity on the right.

    [tex]lny =\mathop {\lim }\limits_{n \to \infty } n ln(1+\frac{x}{n})[/tex]

    Now to simplify and move forward, we must do some algebraic manipulation.

    (Multiply by 1, which in this case we represent with [tex]\frac{\frac{1}{n}}{\frac{1}{n}}[/tex])

    [tex]lny =\mathop {\lim }\limits_{n \to \infty }n ln(1+\frac{x}{n})*\frac{\frac{1}{n}}{\frac{1}{n}}[/tex]

    As you can see, this cancels out the n in the numerator, AND gives us a denominator.

    [tex]lny =\mathop {\lim }\limits_{n \to \infty } \frac{ln(1+\frac{x}{n})}{\frac{1}{n}}[/tex]

    At this point we would re-evaluate the limit, and it would return 0/0 (plug in infinity, it will give 0/0).

    So, we use L'Hopital's Limit Rule to take the derivative of the numerator and denominator, respectively.
    Afterwords you should get,

    [tex]x*\frac{\frac{n^{-2}}{1+\frac{1}{n}}}{n^{-2}}[/tex] Which you can see would cancel out the [tex]n^-2[/tex]'s from the expression. Which would yield:

    [tex]lny =\mathop {\lim }\limits_{n \to \infty }\frac{x}{1+\frac{1}{n}}[/tex]

    Evaluate the limit of the right side, and you'd get [tex]\frac{x}{1+0}[/tex] or simply [tex]x[/tex]

    So you end up with [tex]lny = x[/tex] after which you can exponentiate both sides, and this gives you your final answer!

    [tex]y = e^x[/tex]

    And since y = both your original limit and your new expression e^x...

    [tex]\mathop{\lim }\limits_{n \to \infty } (1+\frac{x}{n})^{n} = e^x[/tex]

    Now there's a few steps taken out of the explanation, such as actually taking the derivatives of L'Hopital's rule... but I would assume you could perform those by yourself.

    I hope this helps!
     
    Last edited: Mar 31, 2009
  6. Apr 1, 2009 #5

    HallsofIvy

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    Staff Emeritus
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    What is your definition of "e"?
     
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