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Thread: Limit of the quotient of two sequences

  1. #1
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    Limit of the quotient of two sequences

    Let $\displaystyle a_n,b_n$ be two sequences verifying that $\displaystyle a_n=\int_{\frac1{n+1}}^{\frac1n}\arcsin(nx)\,dx$ and $\displaystyle b_n=\int_{\frac1{n+1}}^{\frac1n}\arctan(nx)\,dx.$ Compute $\displaystyle \frac{a_n}{b_n}$ as $\displaystyle n\longrightarrow\infty.$

    Note: you can't use the MVT.
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  2. #2
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    Quote Originally Posted by Krizalid View Post
    Let $\displaystyle a_n,b_n$ be two sequences verifying that $\displaystyle a_n=\int_{\frac1{n+1}}^{\frac1n}\arcsin(nx)\,dx$ and $\displaystyle b_n=\int_{\frac1{n+1}}^{\frac1n}\arctan(nx)\,dx.$ Compute $\displaystyle \frac{a_n}{b_n}$ as $\displaystyle n\longrightarrow\infty.$

    Note: you can't use the MVT.
    You can get the result easily using L'Hospital's Rule (of course, do a simple transformation first), which is 2.
    Last edited by mr fantastic; Sep 18th 2009 at 08:44 AM. Reason: Restored original reply
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  3. #3
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    You said it, easily, and yes, that's the answer.

    But that's quite boring!! That's why I posted the problem here! Try it without those things.
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    Quote Originally Posted by Krizalid View Post
    You said it, easily, and yes, that's the answer.

    But that's quite boring!! That's why I posted the problem here! Try it without those things.
    Are you going to use something simple like the squeeze theorem or something more complicated? haha
    Last edited by mr fantastic; Sep 18th 2009 at 08:45 AM. Reason: Restored original reply
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  5. #5
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    when $\displaystyle n \to \infty $ ,

    can we say the integral $\displaystyle a_n = \arcsin(n/n) \delta x $ and $\displaystyle b_n = \arctan(n/n) \delta x $

    thus the limit is $\displaystyle \frac{ \arcsin(1)}{\arctan(1)} = 2 $ ?

    In fact , we are able to evaluate those integrals no matter what the up. and low. limit are .
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    Let $\displaystyle f:\left[\frac{1}{n+1},\frac{1}{n}\right]\to\mathbb{R}, \ f(x)=\arcsin(nx)$.

    f is strictly increasing. Then

    $\displaystyle \frac{1}{n+1}\leq x\leq\frac{1}{n}\Rightarrow\arcsin\frac{n}{n+1}\le q\arcsin(nx)\leq\frac{\pi}{2}\Rightarrow$

    $\displaystyle \Rightarrow\frac{1}{n(n+1)}\arcsin\frac{n}{n+1}\le q a_n\leq\frac{\pi}{2n(n+1)}$ (1)

    Let $\displaystyle g:\left[\frac{1}{n+1},\frac{1}{n}\right]\to\mathbb{R}, \ g(x)=\arctan(nx)$.

    g is strictly increasing. Then

    $\displaystyle \frac{1}{n+1}\leq x\leq\frac{1}{n}\Rightarrow\arctan\frac{n}{n+1}\le q\arctan(nx)\leq\frac{\pi}{4}\Rightarrow$

    $\displaystyle \Rightarrow\frac{1}{n(n+1)}\arctan\frac{n}{n+1}\le q b_n\leq\frac{\pi}{4n(n+1)}\Rightarrow$

    $\displaystyle \Rightarrow\frac{4n(n+1)}{\pi}\leq\frac{1}{b_n}\le q\frac{n(n+1)}{\arctan\frac{n}{n+1}}$ (2)

    Multiply inequalities (1) and (2):

    $\displaystyle \frac{4\arcsin\frac{n}{n+1}}{\pi}\leq\frac{a_n}{b_ n}\leq\frac{\pi}{2\arctan\frac{n}{n+1}}$

    Now apply limit as $\displaystyle n\to\infty$ and we get $\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=2$
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    Quote Originally Posted by red_dog View Post
    Let $\displaystyle f:\left[\frac{1}{n+1},\frac{1}{n}\right]\to\mathbb{R}, \ f(x)=\arcsin(nx)$.

    f is strictly increasing. Then

    $\displaystyle \frac{1}{n+1}\leq x\leq\frac{1}{n}\Rightarrow\arcsin\frac{n}{n+1}\le q\arcsin(nx)\leq\frac{\pi}{2}\Rightarrow$

    $\displaystyle \Rightarrow\frac{1}{n(n+1)}\arcsin\frac{n}{n+1}\le q a_n\leq\frac{\pi}{2n(n+1)}$ (1)

    Let $\displaystyle g:\left[\frac{1}{n+1},\frac{1}{n}\right]\to\mathbb{R}, \ g(x)=\arctan(nx)$.

    g is strictly increasing. Then

    $\displaystyle \frac{1}{n+1}\leq x\leq\frac{1}{n}\Rightarrow\arctan\frac{n}{n+1}\le q\arctan(nx)\leq\frac{\pi}{4}\Rightarrow$

    $\displaystyle \Rightarrow\frac{1}{n(n+1)}\arctan\frac{n}{n+1}\le q b_n\leq\frac{\pi}{4n(n+1)}\Rightarrow$

    $\displaystyle \Rightarrow\frac{4n(n+1)}{\pi}\leq\frac{1}{b_n}\le q\frac{n(n+1)}{\arctan\frac{n}{n+1}}$ (2)

    Multiply inequalities (1) and (2):

    $\displaystyle \frac{4\arcsin\frac{n}{n+1}}{\pi}\leq\frac{a_n}{b_ n}\leq\frac{\pi}{2\arctan\frac{n}{n+1}}$

    Now apply limit as $\displaystyle n\to\infty$ and we get $\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=2$
    nice job! essentially the squeeze theorem.
    Last edited by mr fantastic; Sep 18th 2009 at 08:46 AM. Reason: Restored original reply
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  8. #8
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    That's nice, different solutions.

    I still have a different one, I'll post it when I get back home.
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  9. #9
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    I did that in my head...took 2.3333... seconds;
    but I forgot how
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  10. #10
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    As promised:

    Put $\displaystyle a_n=\frac1n\int_{1-\frac1{n+1}}^1f$ and suppose that $\displaystyle f$ is continuous at $\displaystyle x=1.$ We have $\displaystyle n^{2}a_{n}=n\int_{1-\frac{1}{n+1}}^{1}{f}=\frac{n}{n+1}\cdot (n+1)\int_{1-\frac{1}{n+1}}^{1}{f},$ so let's prove that $\displaystyle (n+1)\int_{1-\frac{1}{n+1}}^{1}{f}\longrightarrow f(1)$ as $\displaystyle n\longrightarrow\infty.$

    Note that is $\displaystyle (n+1)\int_{1-\frac{1}{n+1}}^{1}{f}=(n+1)\int_{1-\frac{1}{n+1}}^{1}{f(1)\,dx}+(n+1)\int_{1-\frac{1}{n+1}}^{1}{\big(f(x)-f(1)\big)\,dx},$ and besides given $\displaystyle \epsilon>0$ we have

    $\displaystyle \begin{aligned}\left| (n+1)\int_{1-\frac{1}{n+1}}^{1}{f}-f(1) \right|&\le(n+1)\int_{1-\frac{1}{n+1}}^{1}{\big|f(x)-f(1)\big|\,dx} \\ & <(n+1)\int_{1-\frac{1}{n+1}}^{1}{\epsilon\,dx} \\ & =\epsilon. \quad\blacksquare\end{aligned}$

    Hence $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,n^{2}a_{n}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n+1}\cdot (n+1)\int_{1-\frac{1}{n+1}}^{1}{f}=1\cdot f(1)=f(1).$

    Finally, on $\displaystyle a_n$ put $\displaystyle x\longmapsto\frac xn$ and we'll get $\displaystyle a_{n}=\frac{1}{n}\int_{1-\frac{1}{n+1}}^{1}{\arcsin(x)\,dx},$ in the same fashion for $\displaystyle b_n,$ thus $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{n}}{b_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n^{2}a_{n}}{n^{2}b_{n}}=\frac{\pi }{2}\cdot \frac{4}{\pi }=2.$
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    Define
    $\displaystyle
    F(x)=\int_0^x f(t) \ dt
    $

    $\displaystyle
    f(1)=F'(1)=\lim_{n\to\infty} \frac{F(1)-F(\frac{n}{n+1})}{1-\frac{n}{n+1}}=\lim_{n\to\infty} (n+1)\int_{\frac{n}{n+1}}^{1}{f(x)}\;dx$$\displaystyle =\lim_{n\to\infty}\frac{n+1}{n} \cdot \lim_{n\to\infty} n\int_{\frac{n}{n+1}}^{1}{f(x)}\;dx=\lim_{n\to\inf ty} n\int_{\frac{n}{n+1}}^{1}{f(x)}\;dx
    $
    Last edited by mr fantastic; Sep 18th 2009 at 08:48 AM. Reason: Restored original reply
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  12. #12
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    Summary, the following methods have been used:
    (1) Mean-value Theorem (not allowed here);
    (2) L'Hospital's Rule;
    (3) Squeeze Theorem;
    (4) Definition of Limit;
    (5) Definition of Derivative;
    (6) .....?
    Last edited by mr fantastic; Sep 18th 2009 at 08:48 AM. Reason: Restored original reply
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