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Math Help - Limit Marathon

  1. #91
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here is another one I came up with, find L.

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

    Where 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
    Hint: Let x=\frac{1}{\varphi}
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  2. #92
    Super Member wingless's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Here is another one I came up with, find L.

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

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

    Let x = \frac{1}{t},

    \lim_{t\to {\infty}}<br />
\bigg ( a^{bt+c} + f \bigg(\frac{1}{t}\bigg)+g\bigg(\frac{1}{t}\bigg)+.  ..\bigg )^{\frac{1}{t}}

    \lim_{t\to{\infty}}\bigg(a^{bt+c}\bigg)^{\frac{1}{  t}}\underbrace{\bigg(1+\frac{f\bigg(\frac{1}{t}\bi  gg)}{a^{bt+c}}+\frac{g\bigg(\frac{1}{t}\bigg)}{a^{  bt+c}}\bigg)^{\frac{1}{t}}}_{1^0=1}

    Those fractions will vanish because a^{bx}\succ{f\bigg(\frac{1}{x}\bigg)}\succ{g\bigg(  \frac{1}{x}\bigg)}\succ\cdots

    \lim_{t\to{\infty}}\bigg(a^{bt} a^c\bigg)^{\frac{1}{t}}

    \lim_{t\to{\infty}}a^b a^{\frac{c}{t}}

    L = a^b
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  3. #93
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wingless View Post
    L=\lim_{x\to{0^{+}}}\bigg(a^{\frac{b}{x}+c}+f(x)+g  (x)+...\bigg)^x

    Let x = \frac{1}{t},

    \lim_{t\to {\infty}}<br />
\bigg ( a^{bt+c} + f \bigg(\frac{1}{t}\bigg)+g\bigg(\frac{1}{t}\bigg)+.  ..\bigg )^{\frac{1}{t}}

    \lim_{t\to{\infty}}\bigg(a^{bt+c}\bigg)^{\frac{1}{  t}}\underbrace{\bigg(1+\frac{f\bigg(\frac{1}{t}\bi  gg)}{a^{bt+c}}+\frac{g\bigg(\frac{1}{t}\bigg)}{a^{  bt+c}}\bigg)^{\frac{1}{t}}}_{1^0=1}

    f and g will vanish because a^{bx}\succ{f\bigg(\frac{1}{x}\bigg)}\succ{g\bigg(  \frac{1}{x}\bigg)}\succ\cdots

    \lim_{t\to{\infty}}\bigg(a^{bt} a^c\bigg)^{\frac{1}{t}}

    \lim_{t\to{\infty}}a^b a^{\frac{c}{t}}

    L = a^b
    Almost identical to my solution ...now come on Wingless let's see your limit problem.
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  4. #94
    Rhymes with Orange Chris L T521's Avatar
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    Mathstud, Are we restricted to post limits that have only one variable? Could we post limits for multivariable functions?
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  5. #95
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Mathstud, Are we restricted to post limits that have only one variable? Could we post limits for multivariable functions?
    I really did not intend it to be so, for if you cannot find limits in one variable what hope do you have in two? (You who hasn't answered any single variable limits yet )

    Don't you have any single variable ones?
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  6. #96
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Don't you have any single variable ones?
    The ones I have are too easy... . I'll think of a couple hard ones...
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  7. #97
    Super Member wingless's Avatar
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    Ok then, here is an easy one:

    \lim_{x\to 0^+} \int_x^{2x}\frac{\sin^m t}{t^{m+1}}~dt,~m \in \mathbb{N}^+
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  8. #98
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wingless View Post
    Ok then, here is an easy one:

    \lim_{x\to 0^+} \int_x^{2x}\frac{\sin^m t}{t^{m+1}}~dt,~m \in \mathbb{N}^+
    Hmm...Maybe I am crazy or stupid or a combintation of both, but

    \lim_<br />
{x\to{0^+}}\int_x^{2x}\frac{\sin^m(t)}{t^{m+1}}dt=  \int_0^0\frac{\sin^m(t)}{t^{m+1}}dt=0

    Is there a hidden indetermincay I am missing?
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  9. #99
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    Here's one I puzzled with a while. You all will probably wade through it like Grant through Richmond.

    No restrictions. Use L'Hopital, whatever. I sure don't care.

    Let f(x) and g(x) be polynomials with real coefficients of degree n+1, n

    respectively, where n\geq{0}, and with positive leading

    coefficients A and B. Evaluate

    L=\lim_{x\to\infty}g(x)\int_{0}^{x}e^{f(t)-f(x)}dt

    in terms of A and B and n
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  10. #100
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Here's one I puzzled with a while. You all will probably wade through it like Grant through Richmond.

    No restrictions. Use L'Hopital, whatever. I sure don't care.

    Let f(x) and g(x) be polynomials with real coefficients of degree n+1, n

    respectively, where n\geq{0}, and with positive leading

    coefficients A and B. Evaluate

    L=\lim_{x\to\infty}g(x)\int_{0}^{x}e^{f(t)-f(x)}dt

    in terms of A and B and n
    so is it

    f(x)=ax^{n+1}+bx^{n}+\cdots

    or is it

    f(x)=ax^{n+1}+bx^n?
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  11. #101
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    Quote Originally Posted by wingless View Post
    Ok then, here is an easy one:

    \lim_{x\to 0^+} \int_x^{2x}\frac{\sin^m t}{t^{m+1}}~dt,~m \in \mathbb{N}^+
    i'll solve a general case, i.e. for any positive real numbers \alpha and r < s: \ \lim_{x\to0+} \int_{rx}^{sx} \frac{\sin^{\alpha}t}{t^{\alpha + 1}} \ dt = \ln \left(\frac{s}{r}\right).

    choose x small enough so that 0 < rx < sx < \frac{\pi}{2}. then for any rx < t < sx: \ 0< \sin t < t < \tan t.

    thus: \frac{\cos^{\alpha}(sx)}{t} < \frac{\cos^{\alpha}t}{t}<\frac{\sin^{\alpha}t}{t^{  \alpha + 1}} < \frac{1}{t}. hence: \cos^{\alpha}(sx)\int_{rx}^{sx} \frac{dt}{t} \leq \int_{rx}^{sx} \frac{\sin^{\alpha}t}{t^{\alpha + 1}} \ dt \leq \int_{rx}^{sx} \frac{dt}{t}. that is:

    \cos^{\alpha}(sx) \ln \left(\frac{s}{r} \right) \leq \int_{rx}^{sx} \frac{\sin^{\alpha} t}{t^{\alpha + 1}} \ dt \leq \ln \left(\frac{s}{r} \right). now since \lim_{x\to 0+}\cos^{\alpha}(sx)=1, by the squeeze theorem:

    \lim_{x\to 0+} \int_{rx}^{sx} \frac{\sin^{\alpha} t}{t^{\alpha + 1}} \ dt = \ln \left(\frac{s}{r} \right). \ \ \ \square
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  12. #102
    Super Member wingless's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Hmm...Maybe I am crazy or stupid or a combintation of both, but

    \lim_<br />
{x\to{0^+}}\int_x^{2x}\frac{\sin^m(t)}{t^{m+1}}dt=  \int_0^0\frac{\sin^m(t)}{t^{m+1}}dt=0

    Is there a hidden indetermincay I am missing?
    Yes. Think of the Riemann sum, and try to see the area under the curve as a rectangle. When x approaches 0, the rectangle will have no width and infinite height. So its area is 0\times \infty. You can express the integral as a Riemann sum to see this indeterminacy.
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  13. #103
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    Quote Originally Posted by galactus View Post
    Here's one I puzzled with a while. You all will probably wade through it like Grant through Richmond.

    No restrictions. Use L'Hopital, whatever. I sure don't care.

    Let f(x) and g(x) be polynomials with real coefficients of degree n+1, n

    respectively, where n\geq{0}, and with positive leading

    coefficients A and B. Evaluate

    L=\lim_{x\to\infty}g(x)\int_{0}^{x}e^{f(t)-f(x)}dt

    in terms of A and B and n
    it's very easy! the answer is \frac{B}{(n+1)A}. so let f(x)=Ax^{n+1} + ... , and g(x)=Bx^n + ... .

    now we have: L=\lim_{x\to\infty}g(x)e^{-f(x)} \int_0^x e^{f(t)} \ dt=\lim_{x\to\infty} \frac{\int_0^x e^{f(t)} \ dt}{\frac{e^{f(x)}}{g(x)}}. L'Hopital now gives us:

    L= \lim_{x\to\infty} \frac{(g(x))^2}{f'(x)g(x)-g'(x)}=\lim_{x\to\infty} \frac{B^2x^{2n} + \text{lower degree terms}}{(n+1)ABx^{2n} + \text{lower degree terms}}

    =\frac{B}{(n+1)A}. \ \ \ \ \square
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  14. #104
    Super Member wingless's Avatar
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    Quote Originally Posted by galactus View Post
    Here's one I puzzled with a while. You all will probably wade through it like Grant through Richmond.

    No restrictions. Use L'Hopital, whatever. I sure don't care.

    Let f(x) and g(x) be polynomials with real coefficients of degree n+1, n

    respectively, where n\geq{0}, and with positive leading

    coefficients A and B. Evaluate

    L=\lim_{x\to\infty}g(x)\int_{0}^{x}e^{f(t)-f(x)}dt

    in terms of A and B and n

    f(x)=ax^{n+1}+~...

    g(x)=bx^{n}+~...

    L=\lim_{x\to \infty}g(x)\int_0^x e^{f(t)-f(x)}dt

    L=\lim_{x\to \infty}\frac{g(x)\int_0^x e^{f(t)}dt}{e^{f(x)}}

    Use L'hopital, (I'm not in the mood to do it the pro way, sorry)

    L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]+g(x)\cdot e^{f(x)}}{e^{f(x)}f'(x)}

    L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]}{e^{f(x)}f'(x)}+\frac{g(x)\cdot e^{f(x)}}{e^{f(x)}f'(x)}

    L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]}{e^{f(x)}f'(x)}+\frac{g(x)}{f'(x)}

    L=\lim_{x\to \infty} \frac{g'(x)\cdot [\int_0^x e^{f(t)}dt]}{e^{f(x)}f'(x)}+\frac{b}{a(n+1)}

    Now it can easily be shown that the other limit goes to zero. Simplify \frac{g'(x)}{f'(x)} and use L'hopital. Then only the other fraction is left.

    So,
    L=\frac{b}{a(n+1)}
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  15. #105
    Eater of Worlds
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    Of course, that is it. I looked at it for a little while before it dawned on me to use L'Hopital, then it falls into place pretty easy.
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