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Math Help - Proving Limits

  1. #1
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    Proving Limits

    I need to prove the following statement

    lim
    x---> Infinity

    3x^5-5logex/5e^x-2x^5

    I have

    logex/x^5 --> 0 as x---> Infinity

    not sure about the other bit.

    Also need to prove the following

    lim
    x--->o

    x^2/1-e^x2 = -1

    Can anyone help?
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  2. #2
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    Re: Proving Limits

    Almost impossible to make any sense of that. Please remember, if nothing else, that the Order of Operations still works.

    These are very different animals:

    x^2 / 1 - e^x2

    and

    x^2 / (1 - e^x2)

    Also, what does "e^x2" mean? Maybe e^{x^{2}}? Or something else?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Proving Limits

    Try to multiply top and bottom by e^{-x} and observe what it happens... and also try to learn latex and observe what it happens ...

    Kind regards

    \chi \sigma
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  4. #4
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    Re: Proving Limits

    Thanks. I have a problem with advance settings, I can get the characters up to place them in the tex. Can anyone help?
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  5. #5
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    Re: Proving Limits

    Sorry I do mean e^x^2 and without the brackets.
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Proving Limits

    You have to calculate:
    \lim_{x \to 0} \frac{x^2}{1-e^{x^2}}=-1
    By using l'Hopital's rule we get:
    \lim_{x \to 0} \frac{2x}{-2x\cdot e^{x^2}}=\lim_{x \to 0} \frac{1}{-e^{x^2}}=\frac{1}{-1}=-1
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: Proving Limits

    The first limit:
    \lim_{x\to \infty}\frac{3x^5-\log(e^x)}{5e^x-2x^5}=...?

    Can we assume \log(e^x) is the natural logarithm?
    (I'm not from the USA, so I don't know if they mean the natural logaritm or the logarithm with base 10, because I always use \ln for the natural logarithm)
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  8. #8
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    Re: Proving Limits

    yes.
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  9. #9
    MHF Contributor Siron's Avatar
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    Re: Proving Limits

    Ok, maybe it can be useful to wright:
    5\log(e\cdot x)=5\log(e)+5\log(x)=5+5\log(x)
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  10. #10
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    Re: Proving Limits

    ok, but how do we get to the limit = 0
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  11. #11
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    Re: Proving Limits

    Quote Originally Posted by Arron View Post
    ok, but how do we get to the limit = 0
    Read the replies given to you! You were told how to do this limit in post #6. If you need more help with it, quote the post and say what you still don't understand.
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  12. #12
    MHF Contributor Siron's Avatar
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    Re: Proving Limits

    To prove that:
    \lim_{x \to \infty} \frac{3x^5-5\log(e\cdot x)}{5\cdot e^x-2x^5}=0
    My solution would be:
    \lim_{x\to \infty}\frac{3x^5-5\log(e\cdot x)}{5\cdot e^x-2x^5}=\lim_{x\to \infty} \frac{3-\frac{5\log(e\cdot x)}{x^5}}{\frac{5\cdot e^x}{x^5}-2}

    In general:
    \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x \to a} f(x)}{\lim_{x\to a}g(x)}

    And so:
    \frac{3-\lim_{x\to \infty }\frac{5\log(e\cdot x)}{x^5}}}{\lim_{x\to \infty}\frac{5\cdot e^x}{x^5}-2}}

    We can calculate now the two limits by using l'Hopitals rule:
    \lim_{x\to \infty}\frac{5\log(e\cdot x)}{x^5}=0
    \lim_{x\to \infty}\frac{5\cdot e^x}{x^5}=\infty

    So that means we get:
    \lim_{x \to \infty} \frac{3}{\infty}=0
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