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Math Help - Limit Marathon

  1. #76
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    Find \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.
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  2. #77
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Find \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.
    a^x=e^{\ln(a)x}=1+\ln(a)x+\frac{\ln^2(a)x^2}{2}+..  .

    a^{-x}=1-\ln(a)x+\frac{\ln(a)^2x^2}{2}-...

    a^{x}-a^{-x}=2\ln(a)x+...\sim{2\ln(a)x}

    and then we have

    log_a(a-x)=\frac{\ln(a-x)}{\ln(a)}=\frac{1}{\ln(a)}\bigg[\ln(a)+\ln\bigg(1-\frac{x}{a}\bigg)\bigg]= \frac{1}{\ln(a)}\bigg[\ln(a)-\bigg(\frac{x}{a}\bigg)-\bigg(\frac{x}{a}\bigg)^2-...\bigg]=1-\frac{\bigg(\frac{x}{a}\bigg)}{\ln(a)}-\frac{\bigg(\frac{x}{a}\bigg)^2}{\ln(a)}


    So then



    1-x-\log_a(a-x)=1-x-1+\frac{\bigg(\frac{x}{a}\bigg)}{\ln(a)}+\frac{\bi  gg(\frac{x}{a}\bigg)^2}{\ln(a)}+...


    So from that we can see that

    1-x-\log_a(a-x)\sim\frac{-(a\ln(a)-1)x}{a\ln(a)}

    \therefore\lim_{x\to{0}}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)}\sim\lim_{x\to{0}}\frac{2\ln(a)x}{\frac{-(a\ln(a)-1)x}{a\ln(a)}}=\frac{-2a\ln^2(a)}{a\ln(a)-1}

    Thanks a lot Krizalid! I really enjoyed that one. It was messy, but still very fun.
    Last edited by Mathstud28; June 12th 2008 at 10:05 AM.
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  3. #78
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    Quote Originally Posted by Krizalid View Post
    Find \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.
    Strangely L'Hospital gives a different answer from Mathstud's answer:

    \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)} = \lim_{x\to0}\frac{a^x \ln a+ a^{-x}\ln a}{-1+\frac1{\ln a (a-x)}} = \frac{2a \ln ^2 a}{1 - a\ln a}

    I must have missed something
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  4. #79
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Strangely L'Hospital gives a different answer from Mathstud's answer:

    \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)} = \lim_{x\to0}\frac{a^x \ln a+ a^{-x}\ln a}{-1+\frac1{\ln a (a-x)}} = \frac{2a \ln ^2 a}{1 - a\ln a}

    I must have missed something
    No error! Look at my fourth step about half way in! I typoed a minus to a plus in the natural log, its my fault, I do these things on the computer not on pencil and paper, so when I make a typo it carries through the rest of the calculation

    So yes you are right, my work will have a negative sign differenece
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  5. #80
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here is another easy one

    \lim_{x\to{0}}x^{x^{x^{x^{x^{x^{x^x}}}}}}
    Hint: Think about pairs,
    Last edited by Mathstud28; June 12th 2008 at 02:06 PM.
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  6. #81
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    How 'bout a clean solution to my last problem?
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  7. #82
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    How 'bout a clean solution to my last problem?
    You mean a non-power series solution? I acutally didnt look at this and see anything else but that, but if there is another sort of solution please say so and I will commence the search for it
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  8. #83
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    Quote Originally Posted by Krizalid View Post
    How 'bout a clean solution to my last problem?
    How do you define a clean solution? Without LHospitals, power series?

    Lets use only the definition of a derivative:

    Then lets use the fact that f(x) = a^x \Rightarrow \lim_{x \to 0} \frac{a^x - 1}{x} = f'(0) = \lim_{-x \to 0}\frac{a^{-x} - 1}{-x} = \lim_{x \to 0}\frac{a^{-x} - 1}{-x}

    Now g(x) = \ln x \Rightarrow \lim_{x \to 0} \frac{\ln a - \ln (a-x)}{x} = g'(a) = \frac1{a}

    So \lim_{x \to 0}\frac{a^x - a^{-x}}{1 - x - \log_a (a - x)} = \lim_{x \to 0}\dfrac{\frac{a^x - 1 -(a^{-x} - 1)}{x}}{\frac{\ln a - x\ln a - \ln (a - x)}{x\ln a}}=\lim_{x \to 0} \dfrac{f'(0) - (-f'(0))}{\frac{g'(a)}{\ln a} - 1}


    So \lim_{x \to 0} \dfrac{f'(0) - (-f'(0))}{\frac{g'(a)}{\ln a} - 1} = \frac{2 \ln a}{\frac1{a\ln a} - 1} = \frac{2 a \ln^2 a}{1 - a\ln a}
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  9. #84
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    Popular Problem:

    \lim_{m \to \infty}\int_{0}^{\pi}\frac{\sin t}{1+\cos ^2 mt}\, dt
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  10. #85
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    Quote Originally Posted by Isomorphism View Post

    How do you define a clean solution? Without LHospitals, power series?
    ... and no derivatives, only a solution which involves basic limits and some of algebra.
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  11. #86
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    Quote Originally Posted by Krizalid View Post
    ... and no derivatives, only a solution which involves basic limits and some of algebra.
    Isnt that unfair because every derivative "involves basic limits and some of algebra"?

    I can understand what you are trying to say only if I see your solution perhaps
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  12. #87
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    Quote Originally Posted by Krizalid View Post
    Find \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.
    we only need two basic limits \lim_{x\to0}\frac{a^x - 1}{x}=\ln a and \lim_{x\to0}\frac{\log_a(1 - \frac{x}{a})}{x}=\frac{-1}{a\ln a}, which are direct results of the definition

    \lim_{t\to{\pm \infty}}\left(1+\frac{1}{t}\right)^t=e: for the first limit put a^x - 1 =\frac{1}{t} and for the second limit put \frac{-x}{a}=\frac{1}{t}. now we have:

    f(x)=\frac{a^x - a^{-x}}{1-x-\log_a(a-x)}=-a^{-x}(1+a^x) \times \frac{\frac{a^x -1}{x}}{1+\frac{\log_a(1-\frac{x}{a})}{x}}. thus: \lim_{x\to0}f(x)=(-2) \times \frac{\ln a}{1 - \frac{1}{a\ln a}}=\frac{2a \ln^2 a}{1 - a\ln a}. \ \ \ \square
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  13. #88
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Let: <br />
a_1 \left( x \right) = x,a_2 \left( x \right) = x^x ,...,a_n \left( x \right) = \left. {x^{x^{ {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu<br />
\raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1  mu}} ^x } } } \right\}{\text{ }}x{\text{ appears n times }}<br />
be a sequence of functions

    We'll show that: <br />
\mathop {\lim }\limits_{x \to 0^+} a_n \left( x \right) = \left\{ \begin{gathered}<br />
0{\text{ if }}n \ne \dot 2 \hfill \\<br />
1{\text{ if }}n = \dot 2 \hfill \\ <br />
\end{gathered} \right.<br />

    We have: <br />
\ln \left( {a_n } \right) = a_{n - 1} \cdot \ln \left( x \right)<br />

    The base cases a_1(x) and a_2(x) are quite easy, now consider <br />
n > 2<br />

    <br />
{\text{if }}n \ne \dot 2 \Rightarrow \mathop {\lim }\limits_{x \to 0^+} \ln \left( {a_n } \right) = \mathop {\lim }\limits_{x \to 0^+} \underbrace {a_{n - 1} }_{ \to 1} \cdot \ln \left( x \right) = - \infty <br />

    Thus we must have: <br />
a_n \to 0<br />

    <br />
{\text{if }}n = \dot 2 \Rightarrow \mathop {\lim }\limits_{x \to 0^+} \ln \left( {a_n } \right) = \mathop {\lim }\limits_{x \to 0^+} \underbrace {a_{n - 1} }_{ \to 0^+} \cdot \ln \left( x \right)<br />

    <br />
u = \ln \left( x \right) \Rightarrow a_{n - 1} = e^{u \cdot \overbrace {e^{u \cdot {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu<br />
\raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1  mu}} ^{e^u } } }^{ \to C \in \mathbb{R}^ + }} \mathop \approx \limits_{u \to - \infty } e^{C \cdot u} <br />

    So that: <br />
\mathop {\lim }\limits_{x \to 0^+} \ln \left( {a_n } \right) = \mathop {\lim }\limits_{u \to - \infty } e^{C \cdot u} \cdot u = 0\therefore a_n \to 1<br />

    This completes the induction

    We are asked <br />
\mathop {\lim }\limits_{x \to 0^+} a_8 \left( x \right) = 1<br />
    Good job Paul, here was my solution

    Consider a function

    f(\xi)=\underbrace{\xi^{\xi^{\xi^{\cdots}}}}_{\tex  t{An even number of times}}

    Now let \psi=\xi^{\xi}

    So now we must find where \psi goes as \xi goes to zero

    Which means we must find \lim_{\xi\to{0}}\xi^{\xi}

    Now by letting \xi=\frac{1}{\varphi}

    we get

    \lim_{\varphi\to\infty}\bigg(\frac{1}{\varphi}\big  g)^{\frac{1}{\varphi}}

    Which by the realtion of the Ratio test and Root test is equivalent to

    \lim_{\varphi\to\infty}\frac{\frac{1}{\varphi+1}}{  \frac{1}{\varphi}}=lim_{\varphi\to\infty}\frac{\va  rphi}{\varphi+1}=1

    \therefore\text{ As }\xi\to{0}\Rightarrow\psi\to{1}

    Giving us

    \lim_{\psi\to{1}}\underbrace{\psi^{\psi^{\psi^{\cd  ots}}}}_{\text{Half number of terms}}=1^{1^{1^{\cdots}}}=1

    \therefore\lim_{\xi\to{0}}f(\xi)=1

    And now seeing that this limit falls under the reign of f(\xi)

    The limit is one
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  14. #89
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    Quote Originally Posted by Isomorphism View Post
    Popular Problem:

    \lim_{m \to \infty}\int_{0}^{\pi}\frac{\sin t}{1+\cos ^2 mt}\, dt
    i think you're assuming that m is an integer. the idea of my solution is very simple but it requires

    some writing ... so i wouldn't be surprised if some typos were found in it! ok, first see that:

    (1) \ \ \ \int_{(k-1)\pi}^{k\pi}\frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx = \int_0^{\pi} \frac{\sin(\frac{k\pi - x}{m})}{1+\cos^2x} \ dx. now, in your integral, let mt = x. then:

    I_m=\int_0^{\pi} \frac{\sin t}{1+\cos^2 mt} \ dt=\frac{1}{m}\int_0^{m\pi} \frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx=\frac{1}{m} \sum_{k=1}^m \int_{(k-1)\pi}^{k\pi}\frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx. thus by (1):

    (2) \ \ \ I_m=\frac{1}{m}\sum_{k=1}^m \int_0^{\pi}\frac{\sin(\frac{k\pi - x}{m})}{1+\cos^2x} \ dx =

    \frac{1}{m}\sum_{k=1}^m \sin(\frac{k\pi}{m})\int_0^{\pi} \frac{\cos(\frac{x}{m})}{1+\cos^2x} \ dx -\frac{1}{m}\sum_{k=1}^m\cos(\frac{k\pi}{m})\int_0^  {\pi}\frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx.

    now since 0 \leq \sin(\frac{x}{m}) \leq \frac{x}{m}, and 0 \leq \sin^2(\frac{x}{2m}) \leq \frac{x^2}{4m^2}, we'll have:

    (3) \ \ \ \lim_{m\to\infty} \int_0^{\pi} \frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx = 0, and:

    (4) \ \ \ \lim_{m\to\infty} \int_0^\pi \frac{\cos(\frac{x}{m})}{1+\cos^2x} \ dx=\lim_{m\to\infty} \int_0^{\pi} \frac{1-2\sin^2(\frac{x}{2m})}{1+\cos^2x} \ dx = \int_0^{\pi} \frac{1}{1+\cos^2x} \ dx=\frac{\pi}{\sqrt{2}}.

    it's also clear, by Riemann sum, that:

    (5) \ \ \ \lim_{m\to\infty}\frac{1}{m} \sum_{k=1}^m \sin(\frac{k\pi}{m}) = \frac{1}{\pi}\int_0^{\pi}\sin x \ dx = \frac{2}{\pi}, and:

    (6) \ \ \ \lim_{m\to\infty} \frac{1}{m}\sum_{k=1}^m \cos(\frac{k\pi}{m}) = \frac{1}{\pi}\int_0^{\pi} \cos x \ dx = 0.

    thus from (2), (3), (4), (5), and (6) we have: \lim_{m\to\infty} I_m = \sqrt{2}. \ \ \ \square
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  15. #90
    MHF Contributor Mathstud28's Avatar
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    Here is another one I came up with, find L.

    L=\lim_{x\to{0^{+}}}\bigg(a^{\frac{b}{x}+c}+f(x)+g  (x)+...\bigg)^x

    Where a^{bx}\succ{f\bigg(\frac{1}{x}\bigg)}\succ{g\bigg(  \frac{1}{x}\bigg)}\succ\cdots

    Restrictions: You may not use asymptotic equivalences, rate of increase, or in any other way disregard terms.

    EDIT: typo, also assuming there is no domain conflictions with other functions
    Last edited by Mathstud28; June 13th 2008 at 07:31 PM.
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