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Thread: Trouble finding N in epsilon-N proof

  1. #1
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    Trouble finding N in epsilon-N proof

    I'm trying to prove $\displaystyle lim(\sqrt{n^2+n}-n)=\frac{1}{2}$. I'm driving myself crazy trying to figure out what N should be.

    I have $\displaystyle \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}$
    so $\displaystyle \epsilon >|\frac{1}{2} - \frac{n}{\sqrt{n^2+n}+n}|.$ Okay to drop the abs value,
    so $\displaystyle \frac{\epsilon-\frac{1}{2}}{n} > -\frac{1}{\sqrt{n^2+n}+n}$.
    so $\displaystyle -\frac{\epsilon-\frac{1}{2}}{n} < \frac{1}{\sqrt{n^2+n}+n}$.
    So $\displaystyle -\frac{n}{\epsilon-\frac{1}{2}}-n < \sqrt{n^2+n}$.

    But the left side is undefined at $\displaystyle \epsilon > \frac{1}{2}$, and worse...it could be negative or positive. So I don't think I can proceed, right? I don't think I need to consider cases, so I think I've already done something wrong. I've tried switching the terms in the distance formula, but I end up with a similar problem (left side could be positive or negative, so I can't square it without losing the inequality).

    Thanks,

    Eric.
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  2. #2
    Super Member PaulRS's Avatar
    Joined
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    Quote Originally Posted by Eric Reed View Post
    I'm trying to prove $\displaystyle lim(\sqrt{n^2+n}-n)=\frac{1}{2}$. I'm driving myself crazy trying to figure out what N should be.

    I have $\displaystyle \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}$
    so $\displaystyle \epsilon >|\frac{1}{2} - \frac{n}{\sqrt{n^2+n}+n}|.$ Okay to drop the abs value,
    so $\displaystyle \frac{\epsilon-\frac{1}{2}}{n} > -\frac{1}{\sqrt{n^2+n}+n}$.
    so $\displaystyle -\frac{\epsilon-\frac{1}{2}}{n} < \frac{1}{\sqrt{n^2+n}+n}$.
    So $\displaystyle -\frac{n}{\epsilon-\frac{1}{2}}-n < \sqrt{n^2+n}$.

    But the left side is undefined at $\displaystyle \epsilon > \frac{1}{2}$, and worse...it could be negative or positive. So I don't think I can proceed, right? I don't think I need to consider cases, so I think I've already done something wrong. I've tried switching the terms in the distance formula, but I end up with a similar problem (left side could be positive or negative, so I can't square it without losing the inequality).

    Thanks,

    Eric.
    Note that: $\displaystyle
    \tfrac{n}
    {{\sqrt {n^2 + n} + n}} = \tfrac{{\tfrac{n}
    {n}}}
    {{\tfrac{{\sqrt {n^2 + n} + n}}
    {n}}} = \tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}}
    $

    You have: $\displaystyle
    \left| {\tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}} - \tfrac{1}
    {2}} \right| < \varepsilon \Leftrightarrow - \varepsilon < \tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}} - \tfrac{1}
    {2} < \varepsilon
    $

    $\displaystyle
    1 + \tfrac{1}
    {n} > 1\therefore \sqrt {1 + \tfrac{1}
    {n}} + 1 > 2\therefore \tfrac{1}
    {2} > \tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}}

    $ thus: $\displaystyle
    \tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}} - \tfrac{1}
    {2} < 0 < \varepsilon
    $

    So you only have to work on the LHS: $\displaystyle
    - \varepsilon < \tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}} - \tfrac{1}
    {2}
    $

    $\displaystyle
    - \varepsilon < \tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}} - \tfrac{1}
    {2} \Leftrightarrow \varepsilon > \tfrac{1}
    {2} - \tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}} \Leftrightarrow \tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}} > \tfrac{1}
    {2} - \varepsilon
    $ now this ineq. is obvious if $\displaystyle
    \varepsilon \geqslant \tfrac{1}
    {2}
    $ for the LHS is positive, thus we may assume that: $\displaystyle
    \varepsilon < \tfrac{1}
    {2}
    $

    $\displaystyle
    \tfrac{1}
    {{\sqrt {1 + \tfrac{1}
    {n}} + 1}} > \tfrac{1}
    {2} - \varepsilon \Leftrightarrow \tfrac{1}
    {{\tfrac{1}
    {2} - \varepsilon }} - 1 > \sqrt {1 + \tfrac{1}
    {n}}
    $ this makes sense since $\displaystyle
    \tfrac{1}
    {{\tfrac{1}
    {2} - \varepsilon }} - 1 > \tfrac{1}
    {{\tfrac{1}
    {2} - 0}} - 1 = 1
    $ so: $\displaystyle
    n > \tfrac{1}
    {{\left( {\tfrac{1}
    {{\tfrac{1}
    {2} - \varepsilon }} - 1} \right)^2 - 1}}
    $
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  3. #3
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hello
    Quote Originally Posted by Eric Reed View Post
    I'm trying to prove $\displaystyle lim(\sqrt{n^2+n}-n)=\frac{1}{2}$. I'm driving myself crazy trying to figure out what N should be.

    I have $\displaystyle \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}$
    so $\displaystyle \epsilon >|\frac{1}{2} - \frac{n}{\sqrt{n^2+n}+n}|.$ Okay to drop the abs value,
    so $\displaystyle \frac{\epsilon-\frac{1}{2}}{n} > -\frac{1}{\sqrt{n^2+n}+n}$.
    so $\displaystyle -\frac{\epsilon-\frac{1}{2}}{n} < \frac{1}{\sqrt{n^2+n}+n}$.
    So $\displaystyle -\frac{n}{\epsilon-\frac{1}{2}}-n < \sqrt{n^2+n}$.

    But the left side is undefined at $\displaystyle \epsilon > \frac{1}{2}$, and worse...it could be negative or positive. So I don't think I can proceed, right? I don't think I need to consider cases, so I think I've already done something wrong. I've tried switching the terms in the distance formula, but I end up with a similar problem (left side could be positive or negative, so I can't square it without losing the inequality).
    One can try working with an equality instead of working with an inequality : the function $\displaystyle f:\begin{array}{l} \mathbb{R}^+\mapsto\left[0,\frac12\right]\\x\mapsto \sqrt{x^2+x}-x\end{array}$ (EDIT : see (*)) is an increasing one since it's derivative, which is $\displaystyle f'(x)=\frac{2x+1}{2\sqrt{x^2+x}}-1$, is always positive :

    $\displaystyle f'(x)>0 \Longleftrightarrow 2x+1-2\sqrt{x^2+x}>0 \Longleftrightarrow 4x^2+4x+1>4x^2+4x \Longleftrightarrow 1>0$

    This is interesting because as $\displaystyle \forall x\geq0, \,0\leq f(x)\leq \frac12\, $, if one solves $\displaystyle \varepsilon = \left| \frac{1}{2}-f(x_0)\right|$ for $\displaystyle x_0$ one gets $\displaystyle \varepsilon > \left| \frac{1}{2}-f(n)\right|$ for all $\displaystyle n>x_0$.

    Let's do it : $\displaystyle \varepsilon = \left| \frac{1}{2}-f(x_0)\right|\Longleftrightarrow \varepsilon = \frac{1}{2}-\sqrt{x_0^2+x_0}+x_0\Longleftrightarrow \left(\frac{1}{2}-\varepsilon-x_0\right)^2=\sqrt{x_0^2+x_0}^2 $

    Expanding we get $\displaystyle \left(\frac{1}{2}-\varepsilon\right)^2 -2\left(\frac{1}{2}-\varepsilon\right)x_0+x_0^2=x_0^2+x_0$ thus $\displaystyle x_0=\frac{\left(\frac{1}{2}-\varepsilon\right)^2}{1-2\left(\frac{1}{2}-\varepsilon\right)}=\frac{\left(\frac{1}{2}-\varepsilon\right)^2}{\varepsilon}$

    EDIT : (*) $\displaystyle \sqrt{x^2+x}-x\leq \frac12 \Longleftrightarrow x^2+x \leq \frac14+x+x^2$ which is always true.
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  4. #4
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    Joined
    Jun 2008
    Posts
    6
    Thanks to both! I gotta get to the point that I can see these simplifications right away. flying squirrel -> unfortunately, we don't have derivatives yet so I don't think I'd get credit for this approach. (Edit...heh... I didn't notice your edit, so I think it's fine...thanks!)
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