Novel solution to $\lim_{n \to \infty}\sqrt{n^2 + n} - n$ Solving $\lim_\limits{n \to \infty}\sqrt{n^2 + n} - n\,$ is a classic question from Rudin Chapter 3.  Its standard solution (on this site and in solutions manuals) is to multiply and divide by the conjugate.  Below I present a novel approach, which to me is clearer and more direct, which I request verification and critique of.
Note that by placing this in Chapter 3, Rudin is requesting a solution without L'Hopital's rule.

Solution
Observe that $(n + \frac{1}{2})^2 = n^2 + n + \frac{1}{4}$, so $\sqrt{n^2 + n} = n + \frac{1}{2} + O(\frac{1}{n})$, from which the solution $\frac{1}{2}$ follows immediately.
Or, more explicitly, define $g$ as a function of $n \in \mathbb{N} \to (-1,0)$  such that $2ng + g + g^2 = -\frac{1}{4}$ and $n + \frac{1}{2} + g > 0$. Dividing both sides by $n$ gives $2g + \frac{g}{n} + \frac{g^2}{n} = -\frac{1}{n}$, and since $|g| < 1, \lim_{n \to \infty}g = 0$. Thus, $(n + \frac{1}{2} + g)^2 = n^2 + n$ and $\sqrt{n^2 + n} - n = \frac{1}{2} + g  \to \frac{1}{2}$.

Discussion
I prefer this solution over the conjugate solution for several reasons:
1. It makes clear why the result is true: $\sqrt{n^2 + n} \approx n + \frac{1}{2}$.
The absolute error is fixed ($\frac{1}{4}$), so the relative error for large $n$ vanishes.
2. The solution uses no tricks, but is developed by estimating $\sqrt{n^2 + n}$, a technique at the heart of analysis. An initial estimate is $n$, but its error is $O(1)$, too large.  So we proceed to $n + \frac{1}{2}$, obvious from $(a + b)^2 = a^2 + 2ab + b^2$.
Contrast this with the conjugate solution, which uses a "trick" seemingly pulled out of a hat.  See e.g. Bergman (p.25):

In this problem you can use the ‘‘trick’’ for simplifying such limits from first-year calculus; ask a
friend if you didn’t learn such a trick. Unfortunately, after the first simplification, the ‘‘obvious’’ next step
is really an application of continuity of the square root function, and we can’t talk about continuity until
Chapter 4. So instead...

3. The calculation avoids messy algebraic manipulations.

Questions

*

*Is my proof correct?

*Is it rigorous? My first line simply assumes that to compensate for a constant error, the offset will be $O(\frac{1}{n})$, since it is multiplied by $n$.  I find this convincing but not rigorous.  But I believe my second paragraph, which introduces an explicit $g$ and bounds it, is indeed rigorous.

*Can the writing be improved?

*Do you agree with the advantages I've argued?


Conclusions
Great responses! The main conclusions so far are:

*

*L.F. and Charles Hudgins showed how the solution above assumes, without proof, that $g$ will always be in $(-1, 0)$

*Multiple fixes were suggested; by far the best, IMO, is Will Jagy's idea to add a lower bound

*I simplified this approach, giving a two sentence proof which I feel is clear, rigorous, and pedagogical, and request verification and critique of.

 A: This may not be to your taste, OP, but here's my attempt to write a proof inspired by what you've already done. This is essentially your idea, just written in a slightly more pedagogical way.
In the year 2022, we may easily graph $\sqrt{n^2 + n} - n$ and see that the limit seems to be approaching $\frac{1}{2}$. Inspired by this, we define
$$
r(n) = \sqrt{n^2 + n} - n - \frac{1}{2}
$$
We need only show $\lim_{n \to \infty} r(n) = 0$. Observe
$$
\left(r(n) + n + \frac{1}{2}\right)^2 = n^2 + n
$$
Expanding, we obtain
$$
r(n)^2 + 2nr(n) + r(n) + n^2 + n + \frac{1}{4} = n^2 + n
$$
Then, rearranging
$$
r(n) = -\frac{1}{2n} \left[r(n)^2 + r(n) + \frac{1}{4}\right]
$$
Now we see that it suffices to show that $r(n)$ is bounded. If $r(n)$ is bounded, then the above equation immediately proves $\lim_{n \to \infty} r(n) = 0$. Even better, we get $r(n) = O(\frac{1}{n})$.
How to show that $r(n)$ is bounded? It is obviously bounded below. Here's a trick for the upper bound (discovered by working backward):
$$
n^2 + 2n + 1 > n^2 + n
$$
but then
$$
(n + 1)^2 > n^2 + n
$$
so
$$
n + 1 > \sqrt{n^2 + n}
$$
therefore
$$
\frac{1}{2} > \sqrt{n^2 + n} - n - \frac{1}{2}
$$
which concludes the proof.
A: I like explicit inequalities both above and below something. Gives comfort, I think.
For $n \geq 1,$
$$ n + \frac{1}{2}   - \frac{1}{8n} \; \;  \color{blue}{<} \; \; \; \sqrt  { \; n^2 + n \;} \; \; \; \color{blue}{<} \; \; n + \frac{1}{2} $$
because
$$  \left(  n + \frac{1}{2}   - \frac{1}{8n}     \right)^2  \; = \; n^2 + n - \frac{1}{8n}  +  \frac{1}{64n^2} \; = \; \;  n^2 + n - \;  \frac{8n-1}{64n^2}$$
to do the other side,
$$  \left(  n + \frac{1}{2}   - \frac{1}{8n}   + \frac{1}{16n^2}    \right)^2  \; = \; n^2 + n   +  \; \frac{ \; 20n^2 - 4n + 1}{256n^4}$$
so $n \geq 1$ gives
$$ n + \frac{1}{2}   - \frac{1}{8n} \; \;  \color{blue}{<} \; \; \; \sqrt  { \; n^2 + n \;} \; \; \; \color{blue}{<} \; \;   n + \frac{1}{2}   - \frac{1}{8n}   + \frac{1}{16n^2}  $$
The coefficients must match those in the Taylor series of $\sqrt{1+x},$  but you don't need to know there is such a thing as calculus to find them.
A: Most likely not a "novel". Just wanted to leave it here as I've constructed a method that doesn't use conjugates.
Let $n=\frac 1x$ as $x\to 0^{+}$. We have:
$$
\begin{align}\lim_{n\to \infty}\sqrt {n^2+n}-n&=\lim_{x\to 0^{+}}\sqrt {\frac 1{x^2}+\frac 1x}-\frac 1x\\
&=\lim_{x\to 0^{+}}\frac {\sqrt {x+1}-1}{x}\end{align}
$$
Now I want to escape the L'Hospital's rule.
Therefore, using the substitution $\sqrt {x+1}=u$ as $u\to 1^{+}$, leads to:
$$
\begin{align}\lim_{x\to 0^{+}}\frac {\sqrt {x+1}-1}{x}&=\lim_{u\to 1^{+}}\frac {u-1}{u^2-1}\\
&=\lim_{u\to 1^{+}}\frac {1}{u+1}\\
&=\frac 12.\end{align}
$$

Without factorization, finally  substitute $u=v+1$ as $v\to 0^{+}$, we have:
$$
\begin{align}\lim_{u\to 1^{+}}\frac {u-1}{u^2-1}&=\lim_{v\to 0^{+}}\frac {v}{v^2+2v}\\
&=\lim_{v\to 0^{+}}\frac {1}{v+2}\\
&=\frac 12.\end{align}
$$

But, I always love using the definition of the derivative:
Using by definition of the derivative, we can also conclude that:
$$
\begin{align}\lim_{x\to 0^{+}}\frac {\sqrt {x+1}-1}{x}&=\lim_{x\to 0^{+}}\frac {\sqrt {x+1}-\sqrt 1}{x+1-1}\\
&=\frac {1}{2\sqrt {x}}{\huge{\mid}}_{x=1}\\
&=\frac 12.\end{align}
$$
A: Drawing on Will Jagy's excellent idea to introduce a lower bound, a revised solution is below.  Please verify and critique.
Observe that $(n+\frac 1 2)^2 = n^2 + n + \frac 1 4$, suggesting that $\sqrt{n^2 + n} = n + \frac 1 2 - O(\frac 1 n)$ and the desired limit is $\frac 1 2$.  To prove this, we need a lower bound on $\sqrt{n^2 + n}$.
The quantity $n + \frac 1 2 - \frac 1 n$ is a lower bound, since $(n + \frac 1 2 - \frac 1 n)^2 = n^2 + n - \frac 7 4 - \frac 1 n + \frac 1 {n^2} < n^2 + n$ for $n \geq 1$.  Since $n + \frac 1 2 - n > \sqrt {n^2 +n} - n > n + \frac 1 2 - \frac 1 n - n$, the Squeeze Theorem tells us that $\lim \sqrt {n^2 +n} - n = \frac 1 2.$
A: It’s an old technique to replace $a-b$ with $\frac{a-b}{\frac{a+b}{a+b}}$ or $\frac{a^2-b^2}{a+b}$.
This is especially useful with square roots. $n=\sqrt{n^2},$so you get $\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}=\frac1{\sqrt{1+1/n}+1}$.
That not also gives you a way to the limit, it also lets you evaluate the expression with little rounding error).
Well, you replaced the limit with $$\lim_{n\to\infty}\sqrt{(n+1/2)^2-1/4}-n$$  Which looks quite like $\frac12,$ but needs proving.
A: Rather than answer your numbered subquestions in detail, I suggest here
a simple and clear alternative proof, using what I think is
essentially the same idea. (It occurred to me in the context of a more
complicated recent question,
and I initially posted it only as a
comment
on an answer to a similar question from 2019, but then I was reminded
that you had recently used a similar idea.)
Let $m = n + \frac12,$ and
$$
f(m) = \sqrt{m^2 - \frac14} = \sqrt{n^2 + n}.
$$
Then $0 < f(m) < m,$ and
$$
m(m - f(m)) < (m + f(m))(m - f(m)) = m^2 - f(m)^2 = \frac14,
$$
therefore
$$
0 < m - f(m) <\frac 1{4m},
$$
therefore $m - f(m) \to 0,$ i.e.
$$
\frac12 + n - \sqrt{n^2 + n} \to 0,
$$
which gives the required result.
