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Mega A Level Maths Thread - Mark V Watch

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    (Original post by RoseGatz)
    Got a question I'm a bit stuck on, care to help Zacken or anyone else for that matter? It's more pure stuff :flutter:

    The function \displaystyle \begin{equation*}f\end{equation*  } is defined by

    \displaystyle \begin{equation*}f(x)= \frac{1}{x^3(1+Inx)}\end{equatio  n*} where \displaystyle \begin{equation*}x>0\end{equatio  n*}

    Show that \displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*} as \displaystyle \begin{equation*}x \rightarrow{0}\end{equation*}

    I've managed to show that \displaystyle \begin{equation*}f(x) \rightarrow\infty\end{equation*} instead of showing \displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*}, got no idea where the minus is coming from.
    Impressive \LaTeX!

    \displaystyle \lim_{x \to 0} \ln x = -\infty \Rightarrow \lim_{x\to 0} (1 + \ln x) = -\infty \Rightarrow \lim_{x \to 0} x^3(1 + \ln x) = 0 but in the negative sense, if you get me?

    Look at the graph of \ln x for 0 < x < 1 to see why \ln x diverges to -\infty.

    Although to be fair - if you have a limit that equals either plus or minus infinity, it's pretty much the same thing in that the limit does not exist.
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    (Original post by Zacken)
    Impressive \LaTeX!

    \displaystyle \lim_{x \to 0} \ln x = -\infty \Rightarrow \lim_{x\to 0} (1 + \ln x) = -\infty \Rightarrow \lim_{x \to 0} x^3(1 + \ln x) = 0 but in the negative sense, if you get me?

    Look at the graph of \ln x for 0 < x < 1 to see why \ln x diverges to -\infty
    Haha I know right! Took me a while to get the hang of it.

    We went through this is class earlier today, we knew it was obviously -\infty considering the graph but is this just an assumption we make? Is there no way to explicitly get \displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*} or do we just go off assumption because of the graph?

    (Original post by Zacken)
    Although to be fair - if you have a limit that equals either plus or minus infinity, it's pretty much the same thing in that the limit does not exist.
    :confused:
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    (Original post by RoseGatz)
    Haha I know right! Took me a while to get the hang of it.

    We went through this is class earlier today, we knew it was obviously -\infty considering the graph but is this just an assumption we make? Is there no way to explicitly get \displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*} or do we just go off assumption because of the graph?
    Saying that \lim_{x \to a} f(x) = \infty is a shorthand way of saying that the limit doesn't exist, i.e: that it diverges. Having anything =\infty nearly almost always never makes sense. So fussing about whether it's plus or minus infinity is something that we don't really tend to do because in either case - we're still just saying that the limit doesn't exist.

    If you want some justification for why \lim_{x \to 0} \ln x = -\infty, you could try looking at \lim_{x \to -1} \ln (1 + x) = \lim_{x \to -1} (x - x^2 /2 + x^3/3 - \cdots ), might work - I haven't thought about it properly though, I'm rushing this out.

    Most of the things you do with limits is based on epsilon-delta proofs, that is, the formal condition for limits and once you start working with infinity in there, that falls apart because you can't have something converge to infinity - so at this level, you're reduced to looking at graphs and making assumptions instead of working with things like you would in a real analysis course.

    This: http://math.stackexchange.com/questi...nity-not-exist may shed some light on what I meant.
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    Isn't the x^3 responsible for its divergence to -infinity, since as x approaches 0 ln(x) becomes large and negative therefore making the fraction small and negative?
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    (Original post by drandy76)
    Isn't the x^3 responsible for its divergence to -infinity, since as x approaches 0 ln(x) becomes large and negative therefore making the fraction small and negative?
    x^3 is what causes the divergence to infinity. x^3 \ln x is what causes the divergence to -\infty. The user is enquiring as to the negative sign.
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    (Original post by Zacken)
    x^3 is what causes the divergence to infinity. x^3 \ln x is what causes the divergence to -\infty. The user is enquiring as to the negative sign.
    sorry misread
 
 
 
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