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Math Help - Interesting result of lim k—»0

  1. #1
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    Lightbulb Interesting result of lim k—»0

    I discovered this interesting result while working on another problem.

    \lim_{k \to 0}\frac{e^{kx}-kx-1}{e^k-k-1}=\ ?


    Can you solve it?

    I have the answer; please PM me if you'd like to see it. I'll post it here in a few days. Good luck!

    Cheers,
    ~ Justin
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  2. #2
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    Re: Interesting result of lim k—»0

    Doesn't look particularly difficult or interesting to me. Just apply L'Hopital's rule twice.
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  3. #3
    Super Member ebaines's Avatar
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    Re: Interesting result of lim k—»0

    The limit as k approaches zero is x^2. Take the the second derivative of the numerator and denominator with respect to k, and the limit as k goes to zero is x^2.
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    Re: Interesting result of lim k—»0

    Well, having forgotten all about L'Hôpital's rule, I thought it was interesting. And I think a student taking calculus might have found it interesting. Oh well, ebaines is right:

    \lim_{k \to 0}\frac{e^{kx}-kx-1}{e^k-k-1}=x^2


    I see there's a tough crowd here. My apologies for bringing a butter knife to a gunfight.
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  5. #5
    Super Member ebaines's Avatar
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    Re: Interesting result of lim k—»0

    Quote Originally Posted by jdmc View Post
    I see there's a tough crowd here. My apologies for bringing a butter knife to a gunfight.
    I hope we don't come across as being mean! You laid out a challenge so our natural inclination is see if we can meet it. I do agree that the limit is interesting in that the result was a bit of a surprise, at least for me, so thanks for bringing it up.
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  6. #6
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    Re: Interesting result of lim k—»0

    Ah, come on- yes, the limit is x^2. But please let us see how you got that? Perhaps we were using a cannon when a screwdriver would work!
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  7. #7
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    Re: Interesting result of lim k—»0

    Even though the problem is easy to solve with the use of L'Hôpital's rule, from an intuitive perspective I still find the result interesting.

    While testing an approach to another problem, I started with the basic function:

    f_1(x)=e^x

    I wanted to modify it so that it would satisfy f(0)=f'(0)=0, and came up with:

    f_2(x)=e^x-x-1

    Then I wanted to modify it further so that it would intersect (1,1) , and came up with:

    f_3(x)=\frac{e^x-x-1}{e-1-1}

    Finally, I wanted to add an argument to vary the curvature of the graph while maintaining the preceding requirements, so I changed it to:

    f_4(x)=\frac{e^{kx}-kx-1}{e^k-k-1}

    While playing around with different values of k in a graphing app, I discovered (graphically, rather than analytically) that the function collimated with x^2 when k\to0 . That was completely unexpected, and, I thought, intriguing… regardless of how elementary the analytic proof is.
    Last edited by jdmc; July 30th 2013 at 01:26 PM.
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