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Math Help - Limit where x is a power

  1. #1
    Junior Member beebe's Avatar
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    Limit where x is a power

    So I have the answer to this problem, and I know it's correct, but I want to be sure that the moves I made were legal to get there. Here's the problem:

     \lim_{x\to\infty}e^{x^2-x}

    Here's what I did:

    =\frac{lim_{x\to\infty}e^{x^2}}{lim_{x\to\infty}e^  x} =\frac{e^{lim_{x\to\infty}x^2}}{e^{lim_{x\to\infty  }x}} =\frac{e^1}{e^1}=1

    Were all my moves legal? Specifically, moving the limit sign up into the exponent. Is there a rule that let's me say \lim_{x \to a}B^x=B^{\lim_{x \to a}x} ?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Limit where x is a power

    Quote Originally Posted by beebe View Post
    So I have the answer to this problem, and I know it's correct, but I want to be sure that the moves I made were legal to get there. Here's the problem:

     \lim_{x\to\infty}e^{x^2-x}

    Here's what I did:

    =\frac{lim_{x\to\infty}e^{x^2}}{lim_{x\to\infty}e^  x} =\frac{e^{lim_{x\to\infty}x^2}}{e^{lim_{x\to\infty  }x}} =\frac{e^1}{e^1}=1

    Were all my moves legal? Specifically, moving the limit sign up into the exponent. Is there a rule that let's me say \lim_{x \to a}B^x=B^{\lim_{x \to a}x} ?
    The correct answer is \infty.

    How did you get that \lim_{x\to\infty}x^2=\lim_{x\to\infty}x=1?
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  3. #3
    Junior Member beebe's Avatar
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    Re: Limit where x is a power

    Sorry, I combined multiple problems when I was typing. The infinity symbols should be 1's. Let me try again:

    \lim_{x \to 1}e^{x^2-x} =\frac{lim_{x \to 1}e^{x^2}}{lim_{x \to 1}e^x} =\frac{e^{lim_{x \to 1}x^2}}{e^{lim_{x \to 1}x}} =\frac{e^1}{e^1}=1
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Limit where x is a power

    Quote Originally Posted by beebe View Post
    Sorry, I combined multiple problems when I was typing. The infinity symbols should be 1's. Let me try again:

    \lim_{x \to 1}e^{x^2-x} =\frac{lim_{x \to 1}e^{x^2}}{lim_{x \to 1}e^x} =\frac{e^{lim_{x \to 1}x^2}}{e^{lim_{x \to 1}x}} =\frac{e^1}{e^1}=1
    That's right.

    An easier way would be \lim_{x \to 1}e^{x^2-x}=e^{\lim_{x \to 1}x^2-x}=e^0=1
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  5. #5
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    Re: Limit where x is a power

    Quote Originally Posted by beebe View Post
    Sorry, I combined multiple problems when I was typing. The infinity symbols should be 1's. Let me try again:
    \lim_{x \to 1}e^{x^2-x} =\frac{lim_{x \to 1}e^{x^2}}{lim_{x \to 1}e^x} =\frac{e^{lim_{x \to 1}x^2}}{e^{lim_{x \to 1}x}} =\frac{e^1}{e^1}=1
    Well you chanced the question!
    You did not have to do that much because e^0=1
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