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Thread: Limit Marathon

  1. #31
    Eater of Worlds
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    Yep. That's it. Good job. It is easy if you spot it. Those are the fun ones.
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  2. #32
    MHF Contributor Mathstud28's Avatar
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    Here is one, have fun with the simplicity

    $\displaystyle \lim_{\psi\to{0^{+}}}\frac{\sin(\sqrt{\psi})\tan(e ^{10\psi}-1)}{\sqrt{15\psi^8-17\psi^3+64\psi}}$
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  3. #33
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by galactus View Post
    $\displaystyle \lim_{n\to\rightarrow{\infty}}\frac{1}{n}\sum_{j=1 }^{n}\sum_{k=1}^{n}\frac{j}{j^{2}+k^{2}}$
    It would seem that this is 0 by a "power counting" argument. But I am having some trouble justifying it Mathematically.

    -Dan
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  4. #34
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here is one, have fun with the simplicity

    $\displaystyle \lim_{\psi\to{0^{+}}}\frac{\sin(\sqrt{\psi})\tan(e ^{10\psi}-1)}{\sqrt{15\psi^8-17\psi^3+64\psi}}$
    hint
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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  5. #35
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here is one, have fun with the simplicity

    $\displaystyle \lim_{\psi\to{0^{+}}}\frac{\sin(\sqrt{\psi})\tan(e ^{10\psi}-1)}{\sqrt{15\psi^8-17\psi^3+64\psi}}$
    This actually turns out to be really easy

    Let $\displaystyle u(0)=0$

    Then $\displaystyle \lim_{x\to{0}}\frac{\tan(u(x))}{u(x)}=1$

    This can be done by making the sub with $\displaystyle \psi=u(x)$

    $\displaystyle \therefore\tan(u(x))\sim{u(x)}$ if $\displaystyle u(0)$

    and the sine case was listed earlier, the other case can be seen before which states that if

    a polynomial approaches zero then it is ~ to its lowest powered x term

    $\displaystyle \therefore\sqrt{15x^8-17x^3+64x}\sim{8\sqrt{x}}$

    $\displaystyle \therefore\lim_{\psi\to{0^{+}}}\frac{\sin(\sqrt{\p si})\tan(e^{10\psi}-1)}{\sqrt{15\psi^8-17\psi^3+64\psi}}\sim\lim_{\psi\to{0^{+}}}\frac{\s qrt{\psi}(e^{10\psi}-1)}{8\sqrt{\psi}}=0$
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  6. #36
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    Quote Originally Posted by galactus View Post
    $\displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{n}\sum_{k= 1}^{n}\frac{j}{j^{2}+k^{2}}$
    nice problem! the sum is a double Riemann sum of $\displaystyle f(x,y)=\frac{x}{x^2+y^2}$ on the unit square:

    $\displaystyle \lim_{n\to\infty} \frac{1}{n}\sum_{j=1}^n \sum_{k=1}^n \frac{j}{j^2 + k^2}= \lim_{n\to\infty} \frac{1}{n^2} \sum_{j=1}^n\sum_{k=1}^n \frac{\frac{j}{n}}{(\frac{j}{n})^2 + (\frac{k}{n})^2}=\int_0^1 \int_0^1 \frac{x}{x^2 + y^2} \ dx \ dy$

    $\displaystyle =\frac{1}{2}\int_0^1 (\ln(y^2 + 1) - 2\ln y) \ dy=\frac{\pi}{4} + \frac{\ln 2}{2}. \ \ \ \square$
    Last edited by NonCommAlg; Jun 10th 2008 at 02:47 PM.
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  7. #37
    MHF Contributor Mathstud28's Avatar
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    Here are two simple ones


    $\displaystyle \lim_{x\to{0}}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}$

    and

    $\displaystyle \lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}$
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  8. #38
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here are two simple ones


    $\displaystyle \lim_{x\to{0}}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}$

    and

    $\displaystyle \lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}$

    Well I am sure there is a better way, but based one mine

    Hint: make a substitution

    Hint:conjugate
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  9. #39
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here are two simple ones


    $\displaystyle \lim_{x\to{0}}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}$

    and

    $\displaystyle \lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}$
    For the first one

    $\displaystyle \lim_{x\to{0}}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}$


    First I let

    $\displaystyle L=\lim_{x\to\infty}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}\Rightarrow{L^2}=\lim_{x\to\infty}\b igg[e^{\frac{5}{x}}-3x\bigg]^{x}$


    I then made the sub, let $\displaystyle \psi=\frac{1}{x}$

    so as $\displaystyle x\to{0}\Rightarrow\psi\to\infty$

    So we have that $\displaystyle x=\frac{1}{\psi}$

    So by subbing we have

    $\displaystyle L^2=\lim_{\psi\to\infty}\bigg[e^{5\psi}-\frac{3}{\psi}\bigg]^{\frac{1}{\psi}}$

    Now there are a few ways this could have been done, but the easiest is the Root/Ratio laws test trick

    $\displaystyle \lim_{\psi\to\infty}\bigg[e^{5\psi}-\frac{3}{\psi}\bigg]^{\frac{1}{\psi}}=\lim_{\psi\to\infty}\frac{e^5\cd ot{e^{5\psi}}-\frac{3}{\psi+1}}{e^{5\psi}-\frac{3}{\psi}}$

    which then by termwise division gives

    $\displaystyle \lim_{\psi\to\infty}\frac{e^5-\frac{3}{(\psi+1)e^{5\psi}}}{1-\frac{3}{\psi\cdot{e^{5\psi}}}}=e^5$

    So $\displaystyle L^2=e^{5}\Rightarrow{L=e^{\frac{5}{2}}}$

    For our second limit

    multiplying by the conjugate gives

    $\displaystyle \lim_{x\to{0}}\frac{\cos(ax)-\cos(bx)}{x^2\bigg[\sqrt{\cos(ax)}+\sqrt{\cos(bx)}\bigg]}$

    Now by breaking it up we get

    $\displaystyle \lim_{x\to{0}}\frac{\cos(ax)-\cos(bx)}{x^2}\cdot\lim_{x\to{0}}\frac{1}{\sqrt{\c os(ax)}+\sqrt{\cos(bx)}}=\lim_{x\to{0}}\frac{\cos( ax)-\cos(bx)}{2x^2}$

    now seeing that

    $\displaystyle \cos(ax)-\cos(bx)=\bigg[1-\frac{a^2x^2}{2}+\frac{a^4x^4}{24}-...\bigg]-\bigg[1-\frac{b^2x^2}{2}+\frac{b^4x^4}{24}-...\bigg]$

    Grouping we get

    $\displaystyle \cos(ax)-\cos(bx)=\bigg[1-1\bigg]+\frac{b^2-a^2}{2}x^2-\frac{b^4+a^2}{24}x^4+...\sim\frac{b^2-a^2}{2}x^2$

    So then we have that

    $\displaystyle \lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}=\lim_{x\to{0}}\frac{\cos(ax)-\cos(bx)}{2x^2}\sim\lim_{x\to{0}}\frac{\frac{b^2-a^2}{2}x^2}{2x^2}=\frac{b^2-a^2}{4}$
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  10. #40
    Moo
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    Hi,

    I've got a question for $\displaystyle \left(e^{\frac 5x}-3x\right)^{\frac x2}$

    Assuming that x tends to 0 from the right (>0), $\displaystyle \frac 5x \to +\infty \implies e^{\frac 5x} \to +\infty$, while $\displaystyle 3x \to 0$.

    So we can neglect 3x with regard to $\displaystyle e^{\frac 5x}$

    Therefore, the limit is $\displaystyle \left(e^{\frac 5x}\right)^{\frac x2}=e^{\frac 52}$


    Also, I think you should be patient, we're not all in the same GMT and we're not as available as you are. So if no one answers within 2 hours, it's not a catastrophe...
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  11. #41
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hi,

    I've got a question for $\displaystyle \left(e^{\frac 5x}-3x\right)^{\frac x2}$

    Assuming that x tends to 0 from the right (>0), $\displaystyle \frac 5x \to +\infty \implies e^{\frac 5x} \to +\infty$, while $\displaystyle 3x \to 0$.

    So we can neglect 3x with regard to $\displaystyle e^{\frac 5x}$

    Therefore, the limit is $\displaystyle \left(e^{\frac 5x}\right)^{\frac x2}=e^{\frac 52}$


    Also, I think you should be patient, we're not all in the same GMT and we're not as available as you are. So if no one answers within 2 hours, it's not a catastrophe...
    Haha thanks for pointing that out, I fixed that, if you look back I just literally stopped. I went and made a sandwich and when I came back I started on the second and didnt finish the first

    As for thse second one

    So if no one answers within 2 hours, it's not a catastrophe...
    Yes it is
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  12. #42
    MHF Contributor Mathstud28's Avatar
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    Ok, here are two more

    This first one I did yesterday on another website, so if you have seen it, dont spoil it..not that its that hard

    $\displaystyle \lim_{x\to{0}}\frac{a^{x^2}-\cos(x)}{\sin(x^2)}$

    and the other is

    $\displaystyle \lim_{n\to\infty}\bigg(\frac{3^n+5^n}{2^n+5^n}\big g)^{n(\sqrt{n+1}-\sqrt{n})}$
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  13. #43
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    $\displaystyle \lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}$
    Just note that: $\displaystyle
    \frac{{\sqrt {\cos \left( {ax} \right)} - \sqrt {\cos \left( {bx} \right)} }}
    {{x^2 }} = - \tfrac{1}
    {{2x}} \cdot \int_b^a {\tfrac{{\sin \left( {x \cdot y} \right)}}
    {{\sqrt {\cos \left( {x \cdot y} \right)} }}dy}
    $

    $\displaystyle
    \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\cos \left( {ax} \right)} - \sqrt {\cos \left( {bx} \right)} }}
    {{x^2 }} = - \tfrac{1}
    {2} \cdot \mathop {\lim }\limits_{x \to 0} \int_b^a {\tfrac{{\sin \left( {x \cdot y} \right)}}
    {{x \cdot \sqrt {\cos \left( {x \cdot y} \right)} }}dy} =
    $$\displaystyle
    - \tfrac{1}
    {2} \cdot \int_b^a {\mathop {\lim }\limits_{x \to 0} \tfrac{{\sin \left( {x \cdot y} \right)}}
    {{x \cdot \sqrt {\cos \left( {x \cdot y} \right)} }}dy} = - \tfrac{1}
    {2} \cdot \int_b^a {y \cdot dy}
    $ (assuming uniform convergence)

    Thus: $\displaystyle
    \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\cos \left( {ax} \right)} - \sqrt {\cos \left( {bx} \right)} }}
    {{x^2 }} = \tfrac{{b^2 - a^2 }}
    {4}
    $
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  14. #44
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    $\displaystyle \lim_{x\to{0}}\frac{a^{x^2}-\cos(x)}{\sin(x^2)}$
    $\displaystyle L=\mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - \cos \left( x \right)}}
    {{\sin \left( {x^2 } \right)}} = \mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - 1 + 1 - \cos \left( x \right)}}
    {{\sin \left( {x^2 } \right)}}
    $

    Now: $\displaystyle
    \sin \left( {x^2 } \right)\mathop \sim \limits_{x \to 0} x^2
    $ and we'll split the limit in 2 parts since both are defined.

    $\displaystyle
    L = \mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - 1}}
    {{x^2 }} + \mathop {\lim }\limits_{x \to 0} \tfrac{{1 - \cos \left( x \right)}}
    {{x^2 }}
    $

    And: $\displaystyle
    \left\{ \begin{gathered}
    a^{x^2 } - 1 = e^{x^2 \ln \left( a \right)} - 1\mathop \sim \limits_{x \to 0} x^2 \ln \left( a \right) \hfill \\
    1 - \cos \left( x \right)\mathop \sim \limits_{x \to 0} \tfrac{{x^2 }}
    {2} \hfill \\
    \end{gathered} \right.
    $

    So we get: $\displaystyle
    L = \ln \left( a \right) + \tfrac{1}
    {2}
    $
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  15. #45
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    $\displaystyle L=\mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - \cos \left( x \right)}}
    {{\sin \left( {x^2 } \right)}} = \mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - 1 + 1 - \cos \left( x \right)}}
    {{\sin \left( {x^2 } \right)}}
    $

    Now: $\displaystyle
    \sin \left( {x^2 } \right)\mathop \sim \limits_{x \to 0} x^2
    $ and we'll split the limit in 2 parts since both are defined.

    $\displaystyle
    L = \mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - 1}}
    {{x^2 }} + \mathop {\lim }\limits_{x \to 0} \tfrac{{1 - \cos \left( x \right)}}
    {{x^2 }}
    $

    And: $\displaystyle
    \left\{ \begin{gathered}
    a^{x^2 } - 1 = e^{x^2 \ln \left( a \right)} - 1\mathop \sim \limits_{x \to 0} x^2 \ln \left( a \right) \hfill \\
    1 - \cos \left( x \right)\mathop \sim \limits_{x \to 0} \tfrac{{x^2 }}
    {2} \hfill \\
    \end{gathered} \right.
    $

    So we get: $\displaystyle
    L = \ln \left( a \right) + \tfrac{1}
    {2}
    $
    I did it very similarly

    $\displaystyle a^{x^2}-\cos(x)=e^{x^2\ln(a)}-\cos(x)=\bigg[1+\ln(a)x^2+\frac{\ln^2(a)x^4}{2}+...\bigg]-\bigg[1-\frac{x^2}{2}+\frac{x^4}{4!}-..\bigg]$

    Combining we get

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

    and $\displaystyle \sin(x^2)=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-...\sim{x^2}$

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