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Math Help - e Limit Problem

  1. #1
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    e Limit Problem

    \lim x \rightarrow \infty

    (e^{x} + x)^{\dfrac{9}{x}}

    (e^{\infty} + \infty)^{\dfrac{9}{\infty}} This seems to come out to 1 as did the limit of (as x approached infinity) x^{\dfrac{4}{x}} But the computer says the answer is NOT 1.
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  2. #2
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    Re: e Limit Problem

    \begin{align*} \lim_{x \to \infty} \left(e^x + x \right)^{9/x} & = \lim_{x \to \infty} \exp\left( \dfrac{9}{x}\ln(e^x + x)\right) \\ & = \exp\left( \lim_{x \to \infty} \dfrac{9\ln(e^x+x)}{x} \right) \\ & = \exp\left(\lim_{x \to \infty} \dfrac{ \tfrac{d}{dx}\left( 9\ln(e^x+x)\right) }{\tfrac{d}{dx}\left(x\right)}\right) \\ & = \exp\left(\lim_{x \to \infty} \dfrac{ \left(\dfrac{ 9(e^x + 1)}{e^x + x} \right) }{1} \right) \\ & = e^9\end{align*}
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  3. #3
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    Re: e Limit Problem

    But how come it didn't evaluate to 1 from the get go? Why did we proceed to the derivative and then plugin?

    What is e^{\infty} ?
    Last edited by Jason76; November 4th 2013 at 11:46 PM.
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    Re: e Limit Problem

    Slip eternal,
    You and I must have been working on this problem at the same time. I did it differently


    lim [ex+x]9/x
    x->infinity

    =lim ex(1+(x/ex )]9/x
    x->infinity

    =lim ex9/x(1+(x/ex )]9/x
    x->infinity

    =lim e9(1+(x/ex )]9/x
    x->infinity

    = e9*1

    = e9

    Extra Little Note:
    SlipEternal turned it into a algebraic fraction so she/he could use L'hopital's Rule to solve it.
    L'H˘pital's rule - Wikipedia, the free encyclopedia
    Last edited by Melody2; November 5th 2013 at 05:26 AM.
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  5. #5
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    Re: e Limit Problem

    Quote Originally Posted by Jason76 View Post
    But how come it didn't evaluate to 1 from the get go? Why did we proceed to the derivative and then plugin?

    What is e^{\infty} ?
    It didn't evaluate to 1 from the get go because you cannot plug in infinity. Infinity is not a number. \exp(f(x)) is another way of writing e^{f(x)}. I used that notation because I wanted to be clear what was in the exponent of e. So, I did e^{\ln(\text{original expression})}. That was how I started. Then, using the properties of natural log, the exponent becomes a coefficient. Since e^{f(x)} is a composition of continuous functions, I used the limit law that states that if g is a continuous function, then \lim_{x \to a} g(f(x)) = g\left(\lim_{x \to a}f(x)\right). That's how I brought the limit inside the \exp(\cdots) function. Then, on line 2, the expression \dfrac{9\ln(e^x + x)}{x} is approaching \dfrac{\infty}{\infty} as x \to \infty, so we can apply L'Hospital's Rule. I took the derivative of the top and the derivative of the bottom. Then, I skipped a step. We had \dfrac{9(e^x+1)}{e^x+x} = \dfrac{9+\tfrac{1}{e^x}}{1+\tfrac{x}{e^x}}. As x \to \infty, \tfrac{1}{e^x}\to 0 and \tfrac{x}{e^x} \to 0. So, \exp\left(\lim_{x\to \infty} \dfrac{9+\tfrac{1}{e^x}}{1+\tfrac{x}{e^x}} \right) = \exp(9) = e^9.

    To recap, e^\infty is not defined because \infty is not a number. So, instead, I altered the form so that I could apply L'Hospital's Rule.
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  6. #6
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    Re: e Limit Problem

    lim [ex+x]9/x
    x->infinity

    I have a slightly different explanation of why you can't just substitute in infinity.

    ex+x is infinitely large as x approaches infinity.

    9/x is infinitely small as x approaches infinity.

    They are not approaching infinity or 0 at the same rate. You have to change the form or it will not be valid.

    Also,

    Using your logic I think you might be tempted to say that

    lim [ex]9/x = 1
    x->infinity


    but you can see that it can be transformed into

    lim e9x/x
    x->infinity

    lim e9
    x->infinity

    = e9

    Think it over, I hope that it will help.
    Last edited by Melody2; November 5th 2013 at 06:06 AM.
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  7. #7
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    Re: e Limit Problem

    Quote Originally Posted by Melody2 View Post
    Slip eternal,
    You and I must have been working on this problem at the same time. I did it differently


    lim [ex+x]9/x
    x->infinity

    =lim ex(1+(x/ex )]9/x
    x->infinity

    =lim ex9/x(1+(x/ex )]9/x
    x->infinity

    =lim e9(1+(x/ex )]9/x
    x->infinity

    = e9*1

    = e9

    Extra Little Note:
    SlipEternal turned it into a algebraic fraction so she/he could use L'hopital's Rule to solve it.
    L'H˘pital's rule - Wikipedia, the free encyclopedia
    To clear up the notation (matching brackets and parentheses), here is Melody2's solution again (which does not need L'Hospital's Rule):

    \begin{align*}\lim_{x \to \infty}(e^x+x)^{9/x} & = \lim_{x \to \infty}\left[e^x\left(1+\dfrac{x}{e^x}\right) \right]^{9/x} \\ & = \lim_{x \to \infty}e^{9x/x}\left(1+\dfrac{x}{e^x}\right)^{9/x} \\ & = \lim_{x \to \infty} e^9\left(1+\dfrac{x}{e^x}\right)^{9/x} \\ & = e^9 \cdot 1 \\ & = e^9\end{align*}
    Thanks from Melody2
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  8. #8
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    Re: e Limit Problem

    Thanks SlipEternal,
    how do you write it out neatly like that?
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  9. #9
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    Re: e Limit Problem

    Quote Originally Posted by Melody2 View Post
    Thanks SlipEternal,
    how do you write it out neatly like that?
    Hit "Reply With Quote" on my previous post to see what I entered. I am using LaTeX formatting.
    Thanks from Melody2
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  10. #10
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    Re: e Limit Problem

    I have just down loaded Latex but I haven't tried to use it yet. I have questions but I probably need to play with it first.
    Thanks
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