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Math Help - Limits

  1. #1
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    Limits

    Limits. Thank You.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Limits. Thank You.
    applying L'Hopital's a second time was not valid. you had the limit going to 1/0. for L'Hopital's we must have the limit going to \frac 00 or \frac {\infty}{\infty}
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    Re:

    Suppose I have (#/0)...

    Would the best solution be to take both the right and left hand limits of the # x is approaching...

    Thks,

    -qbkr21
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Suppose I have (#/0)...

    Would the best solution be to take both the right and left hand limits of the # x is approaching...

    Thks,

    -qbkr21
    Look at it this way:

    we use the limit theorem, If \lim f(x) = F and \lim g(x) = G, then \lim f(x)g(x) = \lim f(x) \cdot \lim g(x) = FG

    thus, \lim_{t \to 0} \frac {e^t}{3t^2} = \lim_{t \to 0} \frac {e^t}3 \cdot \lim_{t \to 0} \frac 1{t^2}


    and of course, you can get really formal with this, using sequential approaches to limits, but let's not go there, since they didn't ask us to "prove"
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    Re:

    Re:
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Re:
    well, \frac {\infty}{0^{+}} makes no sense. but, no, \frac {\infty}0 and \frac 0{\infty} are not indeterminate forms. they are "undefined" forms. there's a diference
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    Re:

    Case in Point...


    Thanks,

    -qbkr21
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Case in Point...


    Thanks,

    -qbkr21
    short of doing a proof for this question, i'd do something like this (i hope it's valid ):

    \lim_{x \to 0^+} \frac {\ln x}x = \lim_{x \to 0^+} \frac 1x \cdot \lim_{x \to 0^+} \ln x

    since \lim_{x \to 0^+} \frac 1x = \infty and \lim_{x to 0^+} \ln x = - \infty, that is, one is positive and the other is negative. we have that the products are negative and going to - \infty

    thus \lim_{x \to 0^+} \frac {\ln x}x = - \infty

    i know that's the answer, i don't know if what i did was 100% mathematically valid though, but it should be

    TPH and Krizalid are the ones who are exceptional with limits
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    Re:

    Cool kind of like setting up an irregular integral. I forgot that +INF * -INF = -INF
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Cool kind of like setting up an irregular integral. I forgot that +INF * -INF = -INF
    yeah, that was the reasoning behind it
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  11. #11
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    Quote Originally Posted by Jhevon View Post
    short of doing a proof for this question, i'd do something like this (i hope it's valid ):

    \lim_{x \to 0^+} \frac {\ln x}x = \lim_{x \to 0^+} \frac 1x \cdot \lim_{x \to 0^+} \ln x
    I knew that we can split the limit if both of them by separate exist.

    Quote Originally Posted by Jhevon View Post
    TPH and Krizalid are the ones who are exceptional with limits
    Well, TPH is better than me; and I'm a simple guy inloved of mathematics. I'm not in the college yet (that frustrates me)
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