Page 6 of 7 FirstFirst ... 234567 LastLast
Results 76 to 90 of 105

Thread: Limit Marathon

  1. #76
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Find $\displaystyle \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.$
    Follow Math Help Forum on Facebook and Google+

  2. #77
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Krizalid View Post
    Find $\displaystyle \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.$
    $\displaystyle a^x=e^{\ln(a)x}=1+\ln(a)x+\frac{\ln^2(a)x^2}{2}+.. .$

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

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

    and then we have

    $\displaystyle 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]=$$\displaystyle \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



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


    So from that we can see that

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

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

    Thanks a lot Krizalid! I really enjoyed that one. It was messy, but still very fun.
    Last edited by Mathstud28; Jun 12th 2008 at 10:05 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #78
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by Krizalid View Post
    Find $\displaystyle \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.$
    Strangely L'Hospital gives a different answer from Mathstud's answer:

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

    I must have missed something
    Follow Math Help Forum on Facebook and Google+

  4. #79
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Isomorphism View Post
    Strangely L'Hospital gives a different answer from Mathstud's answer:

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

    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
    Follow Math Help Forum on Facebook and Google+

  5. #80
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Mathstud28 View Post
    Here is another easy one

    $\displaystyle \lim_{x\to{0}}x^{x^{x^{x^{x^{x^{x^x}}}}}}$
    Hint: Think about pairs,
    Last edited by Mathstud28; Jun 12th 2008 at 02:06 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #81
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    How 'bout a clean solution to my last problem?
    Follow Math Help Forum on Facebook and Google+

  7. #82
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    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
    Follow Math Help Forum on Facebook and Google+

  8. #83
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    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 $\displaystyle f(x) = a^x \Rightarrow \lim_{x \to 0} \frac{a^x - 1}{x} = f'(0) = \lim_{-x \to 0}\frac{a^{-x} - 1}{-x} = \lim_{x \to 0}\frac{a^{-x} - 1}{-x}$

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

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


    So $\displaystyle \lim_{x \to 0} \dfrac{f'(0) - (-f'(0))}{\frac{g'(a)}{\ln a} - 1} = \frac{2 \ln a}{\frac1{a\ln a} - 1} = \frac{2 a \ln^2 a}{1 - a\ln a}$
    Follow Math Help Forum on Facebook and Google+

  9. #84
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Popular Problem:

    $\displaystyle \lim_{m \to \infty}\int_{0}^{\pi}\frac{\sin t}{1+\cos ^2 mt}\, dt$
    Follow Math Help Forum on Facebook and Google+

  10. #85
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    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.
    Follow Math Help Forum on Facebook and Google+

  11. #86
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    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
    Follow Math Help Forum on Facebook and Google+

  12. #87
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Krizalid View Post
    Find $\displaystyle \lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.$
    we only need two basic limits $\displaystyle \lim_{x\to0}\frac{a^x - 1}{x}=\ln a$ and $\displaystyle \lim_{x\to0}\frac{\log_a(1 - \frac{x}{a})}{x}=\frac{-1}{a\ln a}$, which are direct results of the definition

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

    $\displaystyle f(x)=\frac{a^x - a^{-x}}{1-x-\log_a(a-x)}=-a^{-x}(1+a^x) \times \frac{\frac{a^x -1}{x}}{1+\frac{\log_a(1-\frac{x}{a})}{x}}.$ thus: $\displaystyle \lim_{x\to0}f(x)=(-2) \times \frac{\ln a}{1 - \frac{1}{a\ln a}}=\frac{2a \ln^2 a}{1 - a\ln a}. \ \ \ \square$
    Follow Math Help Forum on Facebook and Google+

  13. #88
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by PaulRS View Post
    Let: $\displaystyle
    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
    \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1 mu}} ^x } } } \right\}{\text{ }}x{\text{ appears n times }}
    $ be a sequence of functions

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

    We have: $\displaystyle
    \ln \left( {a_n } \right) = a_{n - 1} \cdot \ln \left( x \right)
    $

    The base cases $\displaystyle a_1(x)$ and $\displaystyle a_2(x)$ are quite easy, now consider $\displaystyle
    n > 2
    $

    $\displaystyle
    {\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
    $

    Thus we must have: $\displaystyle
    a_n \to 0
    $

    $\displaystyle
    {\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)
    $

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

    So that: $\displaystyle
    \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
    $

    This completes the induction

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

    Consider a function

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

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

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

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

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

    we get

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

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

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

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

    Giving us

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

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

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

    The limit is one
    Follow Math Help Forum on Facebook and Google+

  14. #89
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Isomorphism View Post
    Popular Problem:

    $\displaystyle \lim_{m \to \infty}\int_{0}^{\pi}\frac{\sin t}{1+\cos ^2 mt}\, dt$
    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:

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

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

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

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

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

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

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

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

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

    thus from (2), (3), (4), (5), and (6) we have: $\displaystyle \lim_{m\to\infty} I_m = \sqrt{2}. \ \ \ \square$
    Follow Math Help Forum on Facebook and Google+

  15. #90
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Here is another one I came up with, find L.

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

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

    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; Jun 13th 2008 at 07:31 PM.
    Follow Math Help Forum on Facebook and Google+

Page 6 of 7 FirstFirst ... 234567 LastLast

Similar Math Help Forum Discussions

  1. Secondary school's competition - marathon.
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: Aug 30th 2010, 04:45 AM
  2. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 3rd 2009, 05:05 PM
  3. DE Marathon
    Posted in the Differential Equations Forum
    Replies: 26
    Last Post: Jul 21st 2008, 10:15 AM
  4. Small limit marathon
    Posted in the Calculus Forum
    Replies: 40
    Last Post: Jul 20th 2008, 09:40 PM
  5. Geometry Marathon
    Posted in the Geometry Forum
    Replies: 49
    Last Post: Jul 16th 2007, 01:44 PM

Search Tags


/mathhelpforum @mathhelpforum