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Math Help - 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
    \lim{n\to\infty}\sqrt{n+2\sqrt{n}}-\sqrt{n}

    2. Use the definition of the limit to prove that
    \lim_{n\to\infty}\frac{3n^2-1}{n+2}=+\infty
    (I try to solve it like this |\frac{3n^2-1}{n+2}-\infty|=\varepsilon but what then?)
    Last edited by CaptainBlack; November 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
    \lim{n\to\infty}\sqrt{n+2\sqrt{n}}-\sqrt{n}

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

    this is the definition for that case right?

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

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

    so, if \alpha \in R, let \beta := sup \left\{1, \alpha \right\}. Then for all n > \beta, we have  3n > n > \alpha.
    Last edited by kalagota; November 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
    \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 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 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)
    \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:

    \frac{{\left( {\sqrt {n + 2\sqrt n } - \sqrt n } \right)\left( {\sqrt {n + 2\sqrt n } + \sqrt n } \right)}}<br />
{{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{n + 2\sqrt n - n}}<br />
{{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{2\sqrt n }}<br />
{{\sqrt {n + 2\sqrt n } + \sqrt n }}.

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

    \frac{{2\sqrt n }}<br />
{{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{2}<br />
{{\sqrt {1 + \dfrac{2}<br />
{{\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:

    \frac{{\left( {\sqrt {n + 2\sqrt n } - \sqrt n } \right)\left( {\sqrt {n + 2\sqrt n } + \sqrt n } \right)}}<br />
{{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{n + 2\sqrt n - n}}<br />
{{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{{2\sqrt n }}<br />
{{\sqrt {n + 2\sqrt n } + \sqrt n }}.

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

    \frac{{2\sqrt n }}<br />
{{\sqrt {n + 2\sqrt n } + \sqrt n }} = \frac{2}<br />
{{\sqrt {1 + \dfrac{2}<br />
{{\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 ?
    \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 ?
    \lim_{n\to\infty}\sqrt[3]{n^3+9}-n
    The trick is, to apply the identity a^3-b^3=(a-b)(a^2+ab+b^2).

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

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

    Now back substitute and yields

    \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. 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.
    n \ge 4 \Rightarrow \quad 1 \le \sqrt[{2n}]{{1 + 2^{ - n}  + 3\left( {.2} \right)^n }} \le \sqrt[{2n}]{3}<br />
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  11. #11
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    Thanks, I try to solve it (to find lim):
    18. 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 lim_{n}n\frac{1}{2\sqrt{n}} = \infty*0 is undefined
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  12. #12
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    Because 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 a_n = \frac {n \sin \left( \frac 1n \right)}{n \sqrt{n} + 2} ?

    if so, divide through the numerator and denominator by n

    you get that 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. a_{n} = \frac{in^2+2}{n^2-n+1}
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