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Math Help - Small limit marathon

  1. #16
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Moo View Post

    ----------------------------------
    2/
    \lim_{x \to \tfrac \pi 2} \quad 4x \tan(2x)-\frac{\pi}{\cos(2x)}


    ----------------------------------
    This one seems way too easy....

    \lim_{x \to \tfrac \pi 2} \quad 4x \tan(2x)-\frac{\pi}{\cos(2x)}=2\pi\cdot 0-\frac{\pi}{-1}=\color{red}\boxed{\pi}



    --Chris
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  2. #17
    Moo
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    @ Mathstud, #15 : it looks complicated, huh ?

    Correct substitution, but what about this approximation : \cos x \sim 1-\frac{x^2}2 (Taylor series, that's what I think about when talking about approximations) when x \sim 0

    t=x-1

    --> \lim_{t \to 0} \frac{\ln \sin(\tfrac \pi 2+\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln \cos(\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln(1-\tfrac 12 \cdot \tfrac{\pi^2 t^2}4)}{t^2} \sim -\frac{\tfrac{\pi^2t^2}{8}}{t^2}=-\frac{\pi^2}{8}


    @ Chris : Yes
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  3. #18
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    @ Mathstud, #15 : it looks complicated, huh ?

    Correct substitution, but what about this approximation : \cos x \sim 1-\frac{x^2}2 (Taylor series, that's what I think about when talking about approximations) when x \sim 0

    t=x-1

    --> \lim_{t \to 0} \frac{\ln \sin(\tfrac \pi 2+\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln \cos(\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln(1-\tfrac 12 \cdot \tfrac{\pi^2 t^2}4)}{t^2} \sim -\frac{\tfrac{\pi^2t^2}{8}}{t^2}=-\frac{\pi^2}{8}
    Dangit! That probably would have been easier...oh well...same result right?
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  4. #19
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    10/
    \lim_{x \to \tfrac \pi 6} \quad \frac{(x-\tfrac \pi 6)^2}{2 \sin(x)-1}

    Let

    t=x+\frac{\pi}{6}

    So we get

    \lim_{t\to{0}}\frac{t^2}{2\sin\left(x+\frac{\pi}{6  }\right)-1}

    Now note that

    \sin\left(x+\frac{\pi}{6}\right)

    =\sin\left(x-\frac{\pi}{3}+\frac{\pi}{2}\right)

    =\cos\left(x-\frac{\pi}{3}\right)

    =\frac{\cos(x)}{2}+\frac{\sqrt{3}\sin(x)}{2}

    so

    2\sin\left(x+\frac{\pi}{6}\right)=\cos(x)+\sqrt{3}  \sin(x)

    So we have

    \lim_{t\to{0}}\frac{t^2}{\cos(t)+\sqrt{3}\sin(t)+1  }=0\quad\blacksquare
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  5. #20
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Let

    t=x+\frac{\pi}{6}

    So we get

    \lim_{t\to{0}}\frac{t^2}{2\sin\left(x+\frac{\pi}{6  }\right)-1}

    Now note that

    \sin\left(x+\frac{\pi}{6}\right)

    =\sin\left(x-\frac{\pi}{3}+\frac{\pi}{2}\right)

    =\cos\left(x-\frac{\pi}{3}\right)

    =\frac{\cos(x)}{2}+\frac{\sqrt{3}\sin(x)}{2}

    so

    2\sin\left(x+\frac{\pi}{6}\right)=\cos(x)+\sqrt{3}  \sin(x)

    So we have

    \lim_{t\to{0}}\frac{t^2}{\cos(t)+\sqrt{3}\sin(t)+1  }=0\quad\blacksquare
    Ahahahaha...you beat me to it. I pretty much had what you had...

    --Chris
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  6. #21
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Ahahahaha...you beat me to it. I pretty much had what you had...

    --Chris
    Quick fingers
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  7. #22
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post

    \lim_{x \to 0} \quad \left(\frac{a^x+b^x}{2}\right)^{\tfrac 1x}

    I think you are missing a criterion here. And can I pull an IMO and say the answer is (won't give it away) by the relation of the Aritmetic mean?
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  8. #23
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    I think you are missing a criterion here. And can I pull an IMO and say the answer is (won't give it away) by the relation of the Aritmetic mean?
    a and b > 0 ?
    i don't know this relation, but i'd say no

    For the previous one, there's a simpler way, but i'll wait a while
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  9. #24
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    a and b > 0 ?
    a>b\quad{a<b}?
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  10. #25
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    a>b\quad{a<b}?
    Nothing mentioned about it, this means that your method won't work (I can see you dividing by a or b xD)
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  11. #26
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Nothing mentioned about it, this means that your method won't work (I can see you dividing by a or b xD)
    Yeah there goes that, and the Geometric Arithmetic mean thing is so nice. but here goes nothing

    L=\lim_{x\to{0}}\left(\frac{a^x+b^x}{2}\right)^{\f  rac{1}{x}}

    \Rightarrow\ln\left(L\right)=\lim_{x\to{0}}\frac{\  ln\left(\frac{a^x+b^x}{2}\right)}{x}

    =\lim_{x\to{0}}\frac{\ln\left(a^xb^x\cdot\frac{b^{-x}+a^{-x}}{2}\right)}{x}

    =\lim_{x\to{0}}\frac{\ln(a^xb^x)}{x}+\lim_{x\to{0}  }\frac{\ln\left(\frac{a^{-x}+b^{-x}}{2}\right)}{x}

    \Rightarrow{L=\ln(ab)+\lim_{x\to{0}}\frac{\ln\left  (\frac{a^{-x}+b^{-x}}{2}\right)}{x}}

    Now for the second limt

    If we let t=-x

    We get

    -\lim_{t\to{0}}\frac{\ln\left(\frac{a^t+b^t}{2}\rig  ht)}{t}=-\ln(L)



    So

    \ln(L)=\ln(ab)-\ln(L)

    \Rightarrow{2\ln(L)=\ln(ab)}

    \Rightarrow{\ln(L)=\ln\left(\sqrt{ab}\right)}

    \Rightarrow{L=\sqrt{ab}}\quad\blacksquare


    Hooray! That was cool!

    [tex]
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  12. #27
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    I think you are missing a criterion here. And can I pull an IMO and say the answer is (won't give it away) by the relation of the Aritmetic mean?
    I think something is missing as well...however, I think I almost have it...but its somewhat of a pain...

    --Chris

    EDIT: As usual...he beats me again...
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  13. #28
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    I think something is missing as well...however, I think I almost have it...but its somewhat of a pain...

    --Chris

    EDIT: As usual...he beats me again...
    Haha...aww..now I feel bad ...sorry
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  14. #29
    Super Member flyingsquirrel's Avatar
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    Hi !
    Quote Originally Posted by Moo View Post
    11/
    \lim_{x \to a} \quad \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}
    Let f(x)=x-a\ln x and g(x)=\sqrt{2ax-x^2}


    f'(x)= 1-\frac{a}{x}

    f''(x)=\frac{a}{x^2}

    g'(x)=\frac{x-a}{\sqrt{2ax-x^2}}

    \begin{aligned}g''(x)&=\frac{\sqrt{2ax-x^2}-(x-a)\cdot \frac{x-a}{\sqrt{2ax-x^2}}}{2ax-x^2}\\ &=\frac{2ax-x^2-x^2+2ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\\ &=\frac{-2x^2+4ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\end{aligned}


     \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=\frac{f(a)-f(x)}{g(a)-g(x)}\underset{a}{\sim} \frac{(x-a) f'(a)+\frac{(x-a)^2}{2}f''(a)}{(x-a) g'(a)+\frac{(x^-a)^2}{2}g''(a)}=\frac{f''(a)}{g''(a)} since f'(a)=g'(a)=0. As f''(a)=-\frac{1}{a} and g''(a)=\frac{2a^2-a^2}{a\times a^2}=\frac{1}{a},

    \boxed{\lim_{x \to a} \quad \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=-\frac{\frac{1}{a}}{\frac{1}{a}}=-1}
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  15. #30
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hi !


    Let f(x)=x-a\ln x and g(x)=\sqrt{2ax-x^2}


    f'(x)= 1-\frac{a}{x}

    f''(x)=\frac{a}{x^2}

    g'(x)=\frac{x-a}{\sqrt{2ax-x^2}}

    \begin{aligned}g''(x)&=\frac{\sqrt{2ax-x^2}-(x-a)\cdot \frac{x-a}{\sqrt{2ax-x^2}}}{2ax-x^2}\\ &=\frac{2ax-x^2-x^2+2ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\\ &=\frac{-2x^2+4ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\end{aligned}


     \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=\frac{f(a)-f(x)}{g(a)-g(x)}\underset{a}{\sim} \frac{(x-a) f'(a)+\frac{(x-a)^2}{2}f''(a)}{(x-a) g'(a)+\frac{(x^-a)^2}{2}g''(a)}=\frac{f''(a)}{g''(a)} since f'(a)=g'(a)=0. As f''(a)=-\frac{1}{a} and g''(a)=\frac{2a^2-a^2}{a\times a^2}=\frac{1}{a},


    \boxed{\lim_{x \to a} \quad \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=-\frac{\frac{1}{a}}{\frac{1}{a}}=-1}
    Awesome! But isn't this kind of using Cauchy's Mean Value Theorem...which is basically L'hopital's?
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