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FP3 Limits, Is My Answer Sufficient ? Watch

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    Q:

    Show that  \displaystyle F(x) \to -\infty \ \text {as} \ x\to 0

     F(x) = \displaystyle \dfrac{1}{x^3(1+\ln x)}

    I go:

     \displaystyle \dfrac{1}{x^3(1+\ln x)} =  \dfrac{1}{x^3+x^3\ln x)}

    As  \displaystyle x\to 0 \ \text {then} \ x^3\to 0 \ \text {and} \ x^3\ln x \to 0-

    Hence as the denominator tends towards  \displaystyle 0- then clearly  \displaystyle F(x)\to -\infty

    Would that suffice as a valid answer ?

    Secondly, and I would highly appreciate it if people could also answer this so that is completes my knowledge in the topic, if we have

     \displaystyle \dfrac{1}{0} then obviously that is undetermined, but if we have

     \displaystyle \dfrac{1}{x} \and\ x\to 0 then can it be said that as

     \displaystyle x\to 0

     \displaystyle \dfrac {1}{x} \to \infty ?

    Am I wright in thinking we can't write

     \displaystyle\lim_{x\to 0} \dfrac {1}{x} = \infty because the limiting value must be finite and so we write

     \displaystyle\lim_{x\to 0} \dfrac {1}{x} \to \infty or

     \displaystyle\dfrac {1}{x} \to \infty as  \displaystyle x\to 0


    Thnx
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    Bump, anyone ?
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    Hi, firstly, it's fine to write that lim x ->0 1/x = infinity, or rather it would be if it were true (consider a limit approaching from the negative numbers (1/-1, 1/-0.1, 1/-0.01, ...)

    You're saved in this limit because you're restricted to x>0 'cause of the lnx.

    Edit: The edits are because I massively misread your post, sorry about that.
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    (Original post by Jodin)
    Hi, firstly, it's fine to write that lim x ->0 1/x = infinity, or rather it would be if it were true (consider a limit approaching from the negative numbers (1/-1, 1/-0.1, 1/-0.01, ...)

    You're saved in this limit because you're restricted to x>0 'cause of the lnx.

    Edit: The edits are because I massively misread your post, sorry about that.
    Right, so is my answer to the first part of the question correct and sufficient ?
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    (Original post by member910132)
    Q:

    Show that  \displaystyle F(x) \to -\infty \ \text {as} \ x\to 0

     F(x) = \displaystyle \dfrac{1}{x^3(1+\ln x)}

    I go:

     \displaystyle \dfrac{1}{x^3(1+\ln x)} =  \dfrac{1}{x^3+x^3\ln x)}

    As  \displaystyle x\to 0 \ \text {then} \ x^3\to 0 \ \text {and} \ x^3\ln x \to 0-

    Hence as the denominator tends towards  \displaystyle 0- then clearly  \displaystyle F(x)\to -\infty

    Would that suffice as a valid answer ?
    No. \displaystyle 0\cdot \infty or \displaystyle 0\cdot (-\infty) are undefined.
    Arranging F(x)
    \displaystyle F(x)=\frac{\frac{1}{x^3}}{1+lnx}
    So if \displaystyle x\rightarrow 0^{+} the form of the limit will be
    like \diaplaystyle \frac{\infty}{\infty}
    Use the L'Hospital rule.

    Secondly, and I would highly appreciate it if people could also answer this so that is completes my knowledge in the topic, if we have

     \displaystyle \dfrac{1}{0} then obviously that is undetermined,
    ..depending on the number set you are using.
    For calculating limit we can use an extension of Reals including infinte ordinals
    that is \displaystyle \mathbb{R} \cup [-\infty, \infty]
    In this set \displaystyle \frac{1}{0^-}=-\infty and
    \displaystyle \frac{1}{0^+}=\infty
    and f.e. \frac{c}{\pm \infty}=0
    but if we have

     \displaystyle \dfrac{1}{x} \and\ x\to 0 then can it be said that as

     \displaystyle x\to 0

     \displaystyle \dfrac {1}{x} \to \infty ?
    AS I noted above if we have
    \displaystyle \frac{1}{x}
    and \isplaystyle x \to 0^+ through the positive reals or
    \displaystyle x \to 0^- through the negative reals
    are two cases.
    For the first \displaystyle \frac{1}{x} \to +\infty
    for the second \displaystyle \frac{1}{x} \to -\infty

    that is the \displaystyle lim_{x \to 0} \frac{1}{x}
    limit is not exists.
    Am I wright in thinking we can't write

     \displaystyle\lim_{x\to 0} \dfrac {1}{x} = \infty because the limiting value must be finite and so we write

     \displaystyle\lim_{x\to 0} \dfrac {1}{x} \to \infty or

     \displaystyle\dfrac {1}{x} \to \infty as  \displaystyle x\to 0


    Thnx
    No. This limit is not exists in this form (in your Q x>0 because of lnx)
 
 
 
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