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Thread: Limits

  1. #1
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    Limits

    Here's the limit that I need to calculate :

    lim x→y (x/x-y)*∫f(t)dt

    where y is a real constant and f(t) is a continuous function. The integral is defined between y and x.

    I haven't calculated limits for a while now, so I don't even know where to start. If someone could help me, that would be appreciated.

    P.S. I don't need the answer, just a starting point
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by yuki267 View Post
    Here's the limit that I need to calculate :

    lim x→y (x/x-y)*∫f(t)dt

    where y is a real constant and f(t) is a continuous function. The integral is defined between y and x.

    I haven't calculated limits for a while now, so I don't even know where to start. If someone could help me, that would be appreciated.

    P.S. I don't need the answer, just a starting point
    Is this what you are trying to say?

    Find

    \lim_{x\to{y}}\left[\frac{x}{x-y}\cdot\int^x_yf(t)dt\right]

    where y is constant?
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  3. #3
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    Yes.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by yuki267 View Post
    Yes.
    Well, that's a pickle.

    It is clear that \int_y^xf(t)dt\to0 as x\to{y}.

    And \frac{x}{x-y}=\frac{1}{1-\frac{y}{x}}\to\infty as x\to{y}.

    Which is strange...

    This screams the second fundamental theorem though, but I don't know what to do. I'm gonna sit back and watch.

    By the way, welcome to MHF.
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  5. #5
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    Thanks a lot for your help. Hopefully someone will be able to figure it out.
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  6. #6
    MHF Contributor Calculus26's Avatar
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    See attachment for my best shot
    Attached Thumbnails Attached Thumbnails Limits-2dftcquotient.jpg  
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  7. #7
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    I forgot to mention that i have to solve this problem without using L'Hopital. Sorry about that, but thanks anyways.
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  8. #8
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    From the mean value theorem there exists a number \alpha \in[x,y] such that \int_{y}^{x}{f(t)dt}=f(\alpha)(x-y). So what happens with \alpha when x->y and use this to find the limit.
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  9. #9
    MHF Contributor Calculus26's Avatar
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    jetta bra -- tak sa mycket

    forgive my spelling I lived in Sverige in 1976-77 and haven't had much practice
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  10. #10
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    Frankly I think that we need to stipulate that x>y.
    By the mean value theorem for integrals of continuous functions  \left( {\exists c_x  \in \left[ {y,x} \right]} \right)\left[ {\frac{{\int_y^x {f(t)} }}{{x - y}} = f\left( {c_x } \right)} \right].
    Now using this notation, the problem reduces to \lim _{~x \to y^ +  } xf\left( {c_x } \right).
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  11. #11
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    Thank you all for helping me with this problem! I forgot all about the mean value theorem.
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  12. #12
    No one in Particular VonNemo19's Avatar
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    So, what - if anything - is wrong with Calculus26's approach? Is it not valid?
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  13. #13
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    Quote Originally Posted by VonNemo19 View Post
    So, what - if anything - is wrong with Calculus26's approach? Is it not valid?
    Frankly I don’t know how to answer your question.
    I do not like having to view an attachment.
    I think that any serious responder should be able to avoid attachments by posting in LaTeX.
    I guess that I just object to using attachments for anything other than graphs or very technical programming problems.
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  14. #14
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Plato View Post
    Frankly I don’t know how to answer your question.
    I do not like having to view an attachment.
    I think that any serious responder should be able to avoid attachments by posting in LaTeX.
    I guess that I just object to using attachments for anything other than graphs or very technical programming problems.
    Ohhh...k. Anyone else?

    It's just that I've been knid of struggling with these kind of limits. I mean, like why one way is good, but not another. The laws of limits seem strong in calculus26's aproach.
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  15. #15
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    Calculus26's approach is valid. I was close using l'Hospital myself and it's of course a good solution.

    Bra lösning Calculus26.
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