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Math Help - Limit Marathon

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Limit Marathon

    Hello everyone, since every other type of marathon has been done on this site and there has been a lull in the activity around here, I propose a limit marathon. Same concept as all the other; after someone completes a limit and gets the OK from the limit's poster, he/she posts their own limit. The restrictions are, no L'hopital's unless stated otherwise (yes jump for joy) and no graphical help. Also, it would be nice if the poster of the limit would provide their solution. For the first one, we will start off with something simple

    Find \lim_{x\to{0}}\frac{\sin(2\sin(3\sin(4\sin(5x)))))  }{x}
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    I'll take a logical shot seen as though i have never done limits.

    \sin (0)~=~0~~,~~a\times0~=~0

    Thus the limit is 0?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Sean12345 View Post
    I'll take a logical shot seen as though i have never done limits.

    [tex]\sin (0)~=~0~~,~~a\times0~=~0[\math]

    Thus the limit is 0?
    The limit is not zero, sorry.

    The reason being that, your direct substitution method would yield \frac{0}{0} which is indeterminate
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  4. #4
    Super Member wingless's Avatar
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    It's obviously 120, using power series.

    \sin x = x - \frac{x^3}{3!}...

    \sin 5x = 5x - \frac{(5x)^3}{3!}...

    4\sin 5x = 4 \left (5x - \frac{(5x)^3}{3!}... \right )

    \sin (4\sin 5x) = \left [ 4 \left (5x - \frac{(5x)^3}{3!}... \right ) \right ] - \frac{\left [ 4 \left (5x - \frac{(5x)^3}{3!}... \right ) \right ]^3}{3!}...

    You can easily see the pattern here. Now we can conclude,

    \sin ( 2 \sin (3 \sin (4 \sin 5x))))) = 120x + P(x)
    and P(x)'s degree is greater than 1.

    So \lim_{x\to 0}\frac{120x + P(x)}{x} = 120.

    I could also show it using sigma notation and it would be more formal, but characters get too little so.. ;p
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wingless View Post
    It's obviously 120, using power series.

    \sin x = x - \frac{x^3}{3!}...

    \sin 5x = 5x - \frac{(5x)^3}{3!}...

    4\sin 5x = 4 \left (5x - \frac{(5x)^3}{3!}... \right )

    \sin (4\sin 5x) = \left [ 4 \left (5x - \frac{(5x)^3}{3!}... \right ) \right ] - \frac{\left [ 4 \left (5x - \frac{(5x)^3}{3!}... \right ) \right ]^3}{3!}...

    You can easily see the pattern here. Now we can conclude,

    \sin ( 2 \sin (3 \sin (4 \sin 5x))))) = 120x + P(x)
    and P(x)'s degree is greater than 1.

    So \lim_{x\to 0}\frac{120x + P(x)}{x} = 120.

    I could also show it using sigma notation and it would be more formal, but characters get too little so.. ;p
    You can propose your limit

    My solution is as follows

    \lim_{x\to{0}}\frac{\sin(u(x))}{u(x)}\sim{u(x)}\te  xt { iff }u(0)=0

    This can be shown by

    \lim_{x\to{0}}\frac{sin(u(x))}{u(x)}

    Let \psi=u(x)

    and since as u(x)\to{0}=0\Rightarrow{\psi\to{0}} by the above statement we have that

    \lim_{\psi\to{0}}\frac{\sin(\psi)}{\psi}=1

    So then we can see that the above asymptotic equivalence is true

    So we can then see

    \sin(2(\sin(3\sin(4\sin(5x))))\sim{2\sin(3\sin(4\s  in(5x)))} \sim{3!\sin(4\sin(5x))}\sim{4!\sin(5x)}\sim{5!x}

    \therefore\lim_{x\to{0}}\frac{\sin(2\sin(3\sin(4\s  in(5x))))}{x}\sim\lim_{x\to{0}}\frac{5!x}{x}=5!
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    My solution is as follows
    (...)
    Ah, I was so sure that you solved this limit using Power Series, as you always do
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wingless View Post
    Ah, I was so sure that you solved this limit using Power Series, as you always do
    Haha, thats always a good assumption, but alas! This time I think it was not the easiest, or most beautful way. Would you care to post a limit, or shall I post another?
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  8. #8
    Super Member wingless's Avatar
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    I can't think of a hard limit at the moment. Go on and post another
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wingless View Post
    I can't think of a hard limit at the moment. Go on and post another
    Ok, this is one that isnt too hard, but I have figured out only one way of doing it that would fit the restrictions of this marathon

    Calculuate

    \lim_{x\to{0}}x^{\sin(x)}

    EDIT: For this one no power series
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    x^{\sin x}=\exp \left[ \sin x\ln x \right]=\exp \left[ \frac{\sin x}{x}\cdot x\ln x \right].

    As x\to0,\,\frac{\sin x}x and x\ln x exist, and the conclusion follows.
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    x^{\sin x}=\exp \left[ \sin x\ln x \right]=\exp \left[ \frac{\sin x}{x}\cdot x\ln x \right].

    As x\to0,\,\frac{\sin x}x and x\ln x exist, and the conclusion follows.
    That is the one I had. I also came up with this one

    \lim_{x\to{0}}x^{\sin(x)}

    Let \sin(x)=\frac{1}{\psi}

    so that as x\to{0}\Rightarrow{\psi\to\infty}

    So we would then have that

    \lim_{x\to{0}}x^{\sin(x)}=\lim_{\psi\to\infty}\arc  sin\bigg(\frac{1}{\psi}\bigg)^{\frac{1}{\psi}}

    Now using the Root test/Ratio test tric we see that

    \lim_{\psi\to\infty}\arcsin\bigg(\frac{1}{\psi}\bi  gg)^{\frac{1}{\psi}}=\lim_{\psi\to\infty}\frac{\ar  csin\bigg(\frac{1}{\psi+1}\bigg)}{\arcsin\bigg(\fr  ac{1}{\psi}\bigg)}

    Now obviously \frac{1}{\psi+1}\sim\frac{1}{\psi}

    \therefore\lim_{\psi\to\infty}\frac{\arcsin\bigg(\  frac{1}{\psi+1}\bigg)}{\arcsin\bigg(\frac{1}{\psi}  \bigg)}\sim\lim_{\psi\to\infty}\frac{\arcsin\bigg(  \frac{1}{\psi}\bigg)}{\arcsin\bigg(\frac{1}{\psi}\  bigg)}=1


    \therefore\lim_{x\to{0}}x^{\sin(x)}=1

    I think this is pretty


    Go ahead and post yours if you wish
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  12. #12
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    Ok, this is one that isnt too hard, but I have figured out only one way of doing it that would fit the restrictions of this marathon

    Calculuate

    \lim_{x\to{0}}x^{\sin(x)}
    Hmmm, let me see...

    =\lim_{x \to 0} e^{\sin x \ln x} \sim_0 e^{x \ln x}

    By the way, did you know that \sin x \sim_0 x comes from a similar form of power series ? o.O

    The limit of x \ln x in 0 is 0, by increasing comparison.

    Therefore, the limit is 1.




    Edit : eh, I'm more late than I imagined...
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hmmm, let me see...

    =\lim_{x \to 0} e^{\sin x \ln x} \sim_0 e^{x \ln x}

    By the way, did you know that \sin x \sim_0 x comes from a similar form of power series ? o.O

    The limit of x \ln x in 0 is 0, by increasing comparison.

    Therefore, the limit is 1.
    Good job! Are you refering to the fact that \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\sim{x}


    Also, the increasing rates thing suffices, and this is not jibe against you for I am sure you know this. But for other observers, but I think it is more apparent to show it this way

    \lim_{x\to{0}}x\ln(x)

    Let x=\frac{1}{\psi}

    then as x\to{0} we have that \psi\to\infty

    \therefore\lim_{x\to{0}}x\ln(x)=\lim_{x\to\infty}\  frac{-\ln(\psi)}{\psi}

    Which then equals zero due to the fact that \ln(\psi)\prec\psi
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  14. #14
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    Good job! Are you refering to the fact that \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\sim{x}
    Oh actually, it's Taylor series... I never noticed because it's not the same spelling in French so it wasn't the same for me.
    But yeah, this looks like it (we learn it with the o(x^n) notation)
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  15. #15
    Super Member wingless's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Good job! Are you refering to the fact that \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\sim{x}
    Since power series are restricted (by Mathstud) for this one, you can simply multiply \sin x \ln x by \frac{x}{x} like Krizalid did above.
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