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Thread: 1. find the limit 2. use the definition of the limit to prove that

  1. #1
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    1. find the limit 2. use the definition of the limit to prove that

    1. Find the limit
    $\displaystyle \lim{n\to\infty}\sqrt{n+2\sqrt{n}}-\sqrt{n}$

    2. Use the definition of the limit to prove that
    $\displaystyle \lim_{n\to\infty}\frac{3n^2-1}{n+2}=+\infty$
    (I try to solve it like this $\displaystyle |\frac{3n^2-1}{n+2}-\infty|=\varepsilon$ but what then?)
    Last edited by CaptainBlack; Nov 3rd 2007 at 10:32 AM.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kawafis44 View Post
    1. Find the limit
    $\displaystyle \lim{n\to\infty}\sqrt{n+2\sqrt{n}}-\sqrt{n}$

    2. Use the definition of the limit to prove that
    $\displaystyle \lim_{n\to\infty}\frac{3n^2-1}{n+2}=+\infty$
    (I try to solve it like this $\displaystyle |\frac{3n^2-1}{n+2}-\infty|=\varepsilon$ but what then?)
    1)
    $\displaystyle \lim_{n\rightarrow\infty} \sqrt{n+2\sqrt{n}}-\sqrt{n} \cdot \frac{\sqrt{n+2\sqrt{n}}+\sqrt{n}}{\sqrt{n+2\sqrt{ n}}+\sqrt{n}}$

    then use L'hospital's Rule..

    2)
    $\displaystyle \lim_{n\rightarrow\infty}\frac{3n^2-1}{n+2}=+\infty$

    this is the definition for that case right?

    $\displaystyle \lim_{x\rightarrow\infty} f(x) = \infty \Longleftrightarrow \forall \alpha \in R, \exists \beta \in R$ such that if $\displaystyle x > \beta \implies f(x) > \alpha.$

    also, $\displaystyle \frac{3n^2-1}{n+2} = 3n - 6 + \frac{11}{n+2}$
    so that,
    $\displaystyle \lim_{n\rightarrow\infty}\frac{3n^2-1}{n+2} = \lim_{n\rightarrow\infty} 3n - \lim_{n\rightarrow\infty}6 + \lim_{n\rightarrow\infty}\frac{11}{n+2}$

    from here, you only need to show that $\displaystyle \lim_{n\rightarrow\infty} 3n = \infty$

    so, if $\displaystyle \alpha \in R$, let $\displaystyle \beta := sup \left\{1, \alpha \right\}$. Then for all $\displaystyle n > \beta$, we have $\displaystyle 3n > n > \alpha$.
    Last edited by kalagota; Nov 3rd 2007 at 06:53 AM.
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  3. #3
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    Quote Originally Posted by kawafis44 View Post
    1. Find the limit
    $\displaystyle \lim_{n\to\infty}\sqrt{n+2\sqrt{n}}-\sqrt{n}$
    For this one, we don't need L'H˘pital's Rule. After multiplyin' the expression, the numerator will be $\displaystyle 2\sqrt n,$ and we can handle the limit easily.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Krizalid View Post
    For this one, we don't need L'H˘pital's Rule. After multiplyin' the expression, the numerator will be $\displaystyle 2\sqrt n,$ and we can handle the limit easily.
    actually, i really don't see it easily.. ▄ can you show it.. thanks.. ▄
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  5. #5
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    Quote Originally Posted by kalagota View Post
    1)
    $\displaystyle \lim_{n\rightarrow\infty} \sqrt{n+2\sqrt{n}}-\sqrt{n} \cdot \frac{\sqrt{n+2\sqrt{n}}+\sqrt{n}}{\sqrt{n+2\sqrt{ n}}+\sqrt{n}}$
    I'll take it from here:

    $\displaystyle \frac{{\left( {\sqrt {n + 2\sqrt n } - \sqrt n } \right)\left( {\sqrt {n + 2\sqrt n } + \sqrt n } \right)}}
    {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{n + 2\sqrt n - n}}
    {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{2\sqrt n }}
    {{\sqrt {n + 2\sqrt n } + \sqrt n }}.$

    Now, divide by $\displaystyle \sqrt n$ top & bottom, so

    $\displaystyle \frac{{2\sqrt n }}
    {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{2}
    {{\sqrt {1 + \dfrac{2}
    {{\sqrt n }}} + 1}}.$

    The conclusion follows.

    --

    By the way, applyin' L'H˘pital's Rule, it's a mess, with all those roots.
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Krizalid View Post
    I'll take it from here:

    $\displaystyle \frac{{\left( {\sqrt {n + 2\sqrt n } - \sqrt n } \right)\left( {\sqrt {n + 2\sqrt n } + \sqrt n } \right)}}
    {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{n + 2\sqrt n - n}}
    {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{2\sqrt n }}
    {{\sqrt {n + 2\sqrt n } + \sqrt n }}.$

    Now, divide by $\displaystyle \sqrt n$ top & bottom, so

    $\displaystyle \frac{{2\sqrt n }}
    {{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{2}
    {{\sqrt {1 + \dfrac{2}
    {{\sqrt n }}} + 1}}.$

    The conclusion follows.

    --

    By the way, applyin' L'H˘pital's Rule, it's a mess, with all those roots.
    ahh limit as the variable approaches infinity, i forgot it.. hahaha, maybe i'm just tired.. thanks!!
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  7. #7
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    ok, thanks very much
    how to find this lim ?
    $\displaystyle \lim_{n\to\infty}\sqrt[3]{n^3+9}-n$
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  8. #8
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    Quote Originally Posted by kawafis44 View Post
    how to find this lim ?
    $\displaystyle \lim_{n\to\infty}\sqrt[3]{n^3+9}-n$
    The trick is, to apply the identity $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2).$

    Let's just focus on $\displaystyle \sqrt[3]{n^3+9}-n.$ Call $\displaystyle u=\sqrt[3]{n^3+9}.$

    From the expression $\displaystyle u-n,$ we'll multiply and divide by $\displaystyle u^2+nu+n^2,$ so $\displaystyle u-n=\frac{u^3-n^3}{u^2+nu+n^2}.$

    Now back substitute and yields

    $\displaystyle \sqrt[3]{n^3+9}\,-\,n=\frac{n^3+9-n^3}{\sqrt[3]{(n^3+9)^2}+n\sqrt[3]{n^3+9}+n^2}=\frac9{\sqrt[3]{(n^3+9)^2}+n\sqrt[3]{n^3+9}+n^2}.$

    From here, clearly, the limit is 0.
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  9. #9
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    Thanks, Krizalid. Now I see that Chile has similar flag to Poland .
    Now I've got problem with something like that:
    24. $\displaystyle a_{n} = \sqrt[2n]{1+2^{-n}+3(0.2)^{n}}$
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  10. #10
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    If you know the following the limit is obvious.
    $\displaystyle n \ge 4 \Rightarrow \quad 1 \le \sqrt[{2n}]{{1 + 2^{ - n} + 3\left( {.2} \right)^n }} \le \sqrt[{2n}]{3}
    $
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  11. #11
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    Thanks, I try to solve it (to find lim):
    18. $\displaystyle a_{n} = \frac{nsin\frac{1}{n}}{n\sqrt{n}+2} = \frac{1cos\frac{1}{n}}{1\sqrt{n}+n\frac{1}{2\sqrt{ n}}}$
    but $\displaystyle lim_{n}n\frac{1}{2\sqrt{n}} = \infty*0$ is undefined
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  12. #12
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    Because $\displaystyle n\sin \left( {\frac{1}{n}} \right) = \frac{{\sin \left( {\frac{1}{n}} \right)}}{{\frac{1}{n}}} \to 0$ you have a bounded numerator.
    So what can be said of the limit?
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  13. #13
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    so how to solve it?
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kawafis44 View Post
    so how to solve it?
    are you talking about $\displaystyle a_n = \frac {n \sin \left( \frac 1n \right)}{n \sqrt{n} + 2}$ ?

    if so, divide through the numerator and denominator by $\displaystyle n$

    you get that $\displaystyle a_n = \frac {\sin \left( \frac 1n \right)}{\sqrt{n} + \frac 2n}$

    now follow Plato's hint. the numerator is bounded, and the denominator will increase without bound
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  15. #15
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    thanks, now limes of sequences with trigonometry functions are more clear for me . but what about complex numbers? i have got example like that to solve:
    30. $\displaystyle a_{n} = \frac{in^2+2}{n^2-n+1}$
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