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Math Help - Limit Marathon

  1. #46
    Super Member wingless's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \lim_{x\to 0}<br />
\frac{{\sqrt {\cos \left( {ax} \right)}  - \sqrt {\cos \left( {bx} \right)} }}<br />
{{x^2 }}
    Another solution:

    \lim_{x\to 0}<br />
\frac{{\sqrt {\cos \left( {ax} \right)}  - \sqrt {\cos \left( {bx} \right)} }}<br />
{{x^2 }}

    \lim_{x\to 0}<br />
\frac{{\sqrt {\cos \left( {ax} \right)}  - \sqrt {\cos \left( {bx} \right)}}}{x^2}\cdot \frac{{\sqrt {\cos \left( {ax} \right)} + \sqrt {\cos \left( {bx} \right)}}} <br />
{{\sqrt {\cos \left( {ax} \right)} + \sqrt {\cos \left( {bx} \right)}}}

    \lim_{x\to 0} \frac{\cos (ax) - \cos (bx)}{x^2 \left ( \sqrt{\cos (ax)} + \sqrt{\cos (bx)}\right )}

    By using a trig identity,

    \lim_{x\to 0} \frac{ -2 \sin \left ( \frac{a x}{2}-\frac{b x}{2} \right ) \sin \left ( \frac{a x}{2}+\frac{b x}{2} \right ) }{x^2 \left ( \sqrt{\cos (ax)} + \sqrt{\cos (bx)}\right )}

    \lim_{x\to 0} \frac{-2}{\sqrt{\cos (ax)} + \sqrt{\cos (bx)}} \cdot \frac{\sin \left ( x \frac{a-b}{2} \right )}{x} \cdot \frac{\sin \left (x \frac{a +b}{2} \right )}{x}

    By evaluating all limits,

    \frac{b^2-a^2}{4}
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  2. #47
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    Quote Originally Posted by Mathstud28 View Post
    Ok, here are two more

    \lim_{n\to\infty}\bigg(\frac{3^n+5^n}{2^n+5^n}\big  g)^{n(\sqrt{n+1}-\sqrt{n})}

    \bigg(\frac{3^n+5^n}{2^n+5^n}\bigg)^{n(\sqrt{n+1}-\sqrt{n})} \le \bigg(\frac{3^n}{5^n}+1\bigg)^{n(\sqrt{n+1}-\sqrt{n})} = \left[\bigg(\frac{3^n}{5^n}+1\bigg)^\frac{5^n}{3^n}\righ  t]^{n\frac{3^n}{5^n}(\sqrt{n+1}-\sqrt{n})}

    So \lim_{n\to\infty} \bigg(\frac{3^n}{5^n}+1\bigg)^{n(\sqrt{n+1}-\sqrt{n})} = \exp\left( \lim_{n\to\infty} n\frac{3^n}{5^n}(\sqrt{n+1}-\sqrt{n})  \right)

    Obviously: \lim_{n\to\infty}n\frac{3^n}{5^n} = 0 and \lim_{n\to\infty}(\sqrt{n+1}-\sqrt{n}) = 0

    Hence: \lim_{n\to\infty} \bigg(\frac{3^n}{5^n}+1\bigg)^{n(\sqrt{n+1}-\sqrt{n})} = 1

    We see that: \bigg(\frac{3^n+5^n}{2^n+5^n}\bigg)^{n(\sqrt{n+1}-\sqrt{n})} \ge 1

    Thus: \lim_{n\to\infty}\bigg(\frac{3^n+5^n}{2^n+5^n}\big  g)^{n(\sqrt{n+1}-\sqrt{n})} = 1

    I have one limit which I find realy hard, I will post it in a moment (have to check it out)
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  3. #48
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    Try this one (without using Stirling formula and L'Hospital):

    \lim_{n \rightarrow \infty} \frac{(n!)^2 2^{2n}}{\sqrt{n} (2n)!}
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  4. #49
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by albi View Post
    Try this one (without using Stirling formula and L'Hospital):

    \lim_{n \rightarrow \infty} \frac{(n!)^2 2^{2n}}{\sqrt{n} (2n)!}
    Hmm...I can see the answer two ways...but ironically those two ways are L'hopital's and Stirlings approximation ...and I am thinking it will have to do with an integral transform...am Ion the right track?
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  5. #50
    Eater of Worlds
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    Take a look at \int_{-\infty}^{\infty}e^{-x^{2}}dx
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  6. #51
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Take a look at \int_{-\infty}^{\infty}e^{-x^{2}}dx
    Well that would correspond

    \lim_{n\to\infty}\frac{(n!)^24^n}{\sqrt{n}(2n)!}=\  frac{(n^n)^2(e^{-n})^24^n(2\pi{n})}{2\sqrt{\pi}n(n^n)^2(e^{-n})^2}=\sqrt{\pi}

    and \int_{-\infty}^{\infty}e^{-x^2}=2\int_0^{\infty}e^{-x^2}=2\cdot\frac{\sqrt{\pi}}{2}=\sqrt{\pi}
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  7. #52
    MHF Contributor Mathstud28's Avatar
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    Try this one without L'hopitals

    \lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\ln(\sin(b\  xi))}
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  8. #53
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Try this one without L'hopitals

    \lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\ln(\sin(b\  xi))}
    Hint: Power series

    (There might be an easier way to do this)
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  9. #54
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Try this one without L'hopitals

    \lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\ln(\sin(b\  xi))}
    Here is my solution

    \ln(\sin(a\xi))=(sin(a\xi)-1)-\frac{(sin(a\xi)-1)^2}{2}+...\sim\sin(a\xi)-1

    and by similar logic

    \ln(\sin(b\xi))\sim\sin(b\xi)-1

    \therefore\lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\  ln(\sin(b\xi))}\sim\lim_{\xi\to{0}}\frac{\sin(a\xi  )-1}{\sin(b\xi)-1}=\frac{0-1}{0-1}=1
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  10. #55
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    Quote Originally Posted by Mathstud28 View Post
    Hmm...I can see the answer two ways...but ironically those two ways are L'hopital's and Stirlings approximation ...and I am thinking it will have to do with an integral transform...am Ion the right track?
    I don't know what you mean by 'integral transform'. But (relativly) simple solution for this limit i have found a while ago. I use some function usually defined by the integral. What is this function? You just have it in your signature :P
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  11. #56
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here is my solution

    \ln(\sin(a\xi))=(sin(a\xi)-1)-\frac{(sin(a\xi)-1)^2}{2}+...\sim\sin(a\xi)-1
    Are you sure?

    Note that \ln(\sin(a\xi))\rightarrow{-\infty} while \sin(a\xi)-1\rightarrow{-1}

    how can \ln(\sin(a\xi))\sim\sin(a\xi)-1 be true?
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  12. #57
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Are you sure?

    Note that \ln(\sin(a\xi))\rightarrow{-\infty} while \sin(a\xi)-1\rightarrow{-1}

    how can \ln(\sin(a\xi))\sim\sin(a\xi)-1 be true?
    Yeah, I realized this, and I was unsure when I posted it, but I thought "Hey if I am wrong someone will correct me" and here you did. So what would your solution be?

    I am actually kind of stumped without L'hopital's. If it is something easy I am going to kick myself

    Actually

    \frac{\ln(\sin(a\xi))}{\ln(\sin(b\xi))}=\ln\bigg(\  sin(a\xi)^{\frac{1}{\ln(\sin(a\xi))}}\bigg)=L\Righ  tarrow e^{L}=\sin(a\xi)^{\frac{1}{\ln(\sin(b\xi)}}

    So then we have that

    If e^{L}=\lim_{\xi\to{0}}\sin(a\xi)^{\frac{1}{\ln(\si  n(b\xi)}}\Rightarrow{e^{-L}=\sin(a\xi)^{\frac{1}{-\ln(\sin(b\xi)}}}

    Now letting \psi=-\ln(\sin(b\xi))\Rightarrow{\xi=\frac{\arcsin(e^{-\psi})}{b}}

    and as x\to{0} \psi\to\infty

    So we have

    \lim_{\psi\to\infty}\bigg[\sin\bigg(\frac{a\arcsin(e^{-\psi})}{b}\bigg)\bigg]^{\frac{1}{\psi}}

    Which by the root/ratio test trick

    we get

    this limit is equal to

    \lim_{\psi\to\infty}\frac{\sin\bigg(\frac{a\arcsin  (e^{-\psi-1})}{b}\bigg)}{\sin\bigg(\frac{a\arcsin(e^{-\psi})}{b}\bigg)}

    And since it can be seen that as x\to\infty that \sin\bigg(\frac{a\arcsin(e^{-\psi-1})}{b}\bigg)\sim\frac{a\arcsin(e^{-\psi-1})}{b}\sim\frac{ae^{-\psi-1}}{b}

    and it can also be said that \sin\bigg(\frac{a\arcsin(e^{-\psi})}{b}\bigg)\sim\frac{ae^{-\psi}}{b}

    So we have that this whole mess above is equivalent to

    \lim_{\psi\to\infty}\frac{e^{-\psi1}}{e^{-\psi}}=\frac{1}{e}

    So now we have

    e^{-L}=\frac{1}{e}\Rightarrow{L=1}


    2th post!
    Last edited by Mathstud28; June 11th 2008 at 01:05 PM.
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  13. #58
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Just note that: <br />
\sin \left( {a \cdot x} \right) = a \cdot x \cdot \left[ {1 + o\left( 1 \right)} \right] \Rightarrow \ln \left[ {\sin \left( {a \cdot x} \right)} \right] = \ln \left( a \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right]<br />
when x\rightarrow{0^+}

    <br />
\mathop {\lim }\limits_{x \to 0^ + } \frac{{\ln \left[ {\sin \left( {a \cdot x} \right)} \right]}}<br />
{{\ln \left[ {\sin \left( {b \cdot x} \right)} \right]}} = \mathop {\lim }\limits_{x \to 0^ + } \frac{{\ln \left( a \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right]}}<br />
{{\ln \left( b \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right]}}<br />

    And: <br />
\frac{{\ln \left( a \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right]}}<br />
{{\ln \left( b \right) + \ln \left( x \right) + \ln \left[ {1 + o\left( 1 \right)} \right]}} = <br /> <br />
<br />
\frac{{\tfrac{{\ln \left( a \right)}}<br />
{{\ln \left( x \right)}} + 1 + \tfrac{{\ln \left[ {1 + o\left( 1 \right)} \right]}}<br />
{{\ln \left( x \right)}}}}<br />
{{\tfrac{{\ln \left( b \right)}}<br />
{{\ln \left( x \right)}} + 1 + \tfrac{{\ln \left[ {1 + o\left( 1 \right)} \right]}}<br />
{{\ln \left( x \right)}}}} \to 1<br />

    <br />
\mathop {\lim }\limits_{x \to 0^ + } \frac{{\ln \left[ {\sin \left( {a \cdot x} \right)} \right]}}<br />
{{\ln \left[ {\sin \left( {b \cdot x} \right)} \right]}} = 1<br />
    I thought of doing it your way, except I would have just said \sin(a\xi)\sim{a\xi} and done simliarly to you, but I do that so much I wanted to think of a more difficult way....you can find such a complicated way above
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  14. #59
    MHF Contributor Mathstud28's Avatar
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    Here is another easy one

    \lim_{x\to{0}}x^{x^{x^{x^{x^{x^{x^x}}}}}}
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  15. #60
    Math Engineering Student
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    Try this one: \lim_{\alpha\to\infty}\frac1{\alpha^2}\int_0^\alph  a\ln(1+e^{x})\,dx.
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