<s>[INST] Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
 [/INST] Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form:
$$
f(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} - x
$$
where
$$
n,k \in \Bbb N \\
a_k \in \Bbb R
$$
Consider the following expansion:
$$
a^n – b^n = (a – b)\left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + 
\cdots + ab^{n – 2} + b^{n – 1}\right) \tag1
$$
Define a function in the form:
$$
g(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)}
$$
Thus $f(x)$ may be rewritten as:
$$
f(x) = g(x) - x
$$
Now using $(1)$ we may rewrite $f(x)$ as:
$$
f(x) = \frac{(g(x))^n-x^n}{\sqrt[n]{(g(x))^{n-1}} +\sqrt[n]{(g(x))^{n-2}}x + \sqrt[n]{(g(x))^{n-3}}x^2 + \cdots + \sqrt[n]{g(x)}x^{n-2} +x^{n-1}}  \tag2
$$
Now taking the limit of $(2)$ one may obtain:
$$
\begin{align*}
&\lim_{x\to\infty}\frac{x^{n-1}a_1 + x^{n-1}a_2+ \cdots + x^{n-1}a_n }{x^{n-1}\left(\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1\right)} \\ 
= &\lim_{x\to\infty}\frac{a_1 + a_2+ \cdots + a_n}{\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1} \\
= &\frac{a_1 + a_2 + \cdots + a_n}{n} \tag3
\end{align*}
$$
Or summarizing:
$$
\boxed{\lim_{x\to\infty}f(x) = \frac{a_1 + a_2 + \cdots + a_n}{n}}
$$

Lets now use that result in your limit:
$$
\sqrt{n^2 + n} - n = \sqrt{n(n+1)} - n = \sqrt{(n+0)(n+1)} - n
$$
Which gives $a_1 = 0$ and $a_2 = 1$ and the highest power is $2$ meaning $n = 2$ in $(3)$. Thus we get:
$$
\boxed{\lim_{n\to\infty}x_n = \frac{a_1 + a_2}{2} = \frac{0 + 1}{2} = {1\over 2}}
$$
As desired.
</s>