# Trouble finding N in epsilon-N proof

• Jul 5th 2008, 11:43 AM
Eric Reed
Trouble finding N in epsilon-N proof
I'm trying to prove $lim(\sqrt{n^2+n}-n)=\frac{1}{2}$. I'm driving myself crazy trying to figure out what N should be.

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

But the left side is undefined at $\epsilon > \frac{1}{2}$, 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.
• Jul 5th 2008, 12:17 PM
PaulRS
Quote:

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

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

But the left side is undefined at $\epsilon > \frac{1}{2}$, 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: $
\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: $
\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
$

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

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

$
- \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 $
\varepsilon \geqslant \tfrac{1}
{2}
$
for the LHS is positive, thus we may assume that: $
\varepsilon < \tfrac{1}
{2}
$

$
\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 $
\tfrac{1}
{{\tfrac{1}
{2} - \varepsilon }} - 1 > \tfrac{1}
{{\tfrac{1}
{2} - 0}} - 1 = 1
$
so: $
n > \tfrac{1}
{{\left( {\tfrac{1}
{{\tfrac{1}
{2} - \varepsilon }} - 1} \right)^2 - 1}}
$
• Jul 5th 2008, 12:37 PM
flyingsquirrel
Hello
Quote:

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

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

But the left side is undefined at $\epsilon > \frac{1}{2}$, 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 $f:\begin{array}{l} \mathbb{R}^+\mapsto\left[0,\frac12\right]\\x\mapsto \sqrt{x^2+x}-x\end{array}$ (EDIT : see (*)) is an increasing one since it's derivative, which is $f'(x)=\frac{2x+1}{2\sqrt{x^2+x}}-1$, is always positive :

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

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

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

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

EDIT : (*) $\sqrt{x^2+x}-x\leq \frac12 \Longleftrightarrow x^2+x \leq \frac14+x+x^2$ which is always true.
• Jul 5th 2008, 12:43 PM
Eric Reed
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!)