Hello

Originally Posted by

**Eric Reed** 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.