<s>[INST] Proving that limit (without using calculus) $(n^2 + n) ^{1/2} - n = \frac{1}{2}$ Prove, Without using Calculus. 
$$\lim_{n->\infty} \left((n^2 + n)^{1/2} - n\right) = \frac{1}{2} $$
I am taking an introduction to real analysis course going through Rudin. 
My work so far, 
By definition, limit of a sequence $p$ has the following property: 
For every $\varepsilon > 0$ There is an integer $N$ s.t. $n \geq N$ implies that $d(p_n, p) < \varepsilon $. 
Let $ \varepsilon  > 0$. Then we need to find an expression for $N$. So: 
$$ \left|\left((N^2 + N)^{1/2} - N\right) - \frac{1}{2} \right| < \varepsilon $$
I am wondering whether this is the right approach?  
 [/INST] Yep it's essentially the right approach. We want to find $N$ so that if $n > N$, then 
$$ \left|\left((n^2 + n)^{1/2} - n\right) - \frac{1}{2} \right| < \varepsilon $$
We know that
$\left((n^2+n)^{1/2} - (n + 1/2)\right)\left((n^2+n)^{1/2} + (n+1/2)\right) = (n^2+n)-(n+1/2)^2$ using a "difference of two squares". Hence given some $n \in \mathbb{N}$
$$\left|\left((n^2 + n)^{1/2} - n\right) - \frac{1}{2} \right| = \frac{\left|(n^2+n)-(n+1/2)^2 \right| }{\left|\left((n^2+n)^{1/2} + (n+1/2)\right)  \right|} = \frac{1/4}{(n^2+n)^{1/2}+n+1/2} < \frac{1}{4n}<\frac{1}{n}$$
We see that if $N = \frac{1}{\varepsilon}$, then for any $n > N$,
$$\left|\left((n^2 + n)^{1/2} - n\right) - \frac{1}{2} \right| <\frac{1}{n} < \frac{1}{N} = \varepsilon$$
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