Justification for Rudin exercise 3.2 
Rudin 3.4. Calculate $\lim\limits_{n \to \infty} \left(\sqrt{n^2 + n} - n\right)$.

The solution to this seems obvious: after enough arithmetic, let $n \to \infty$ and get $\frac{1}{2}$. I am not sure how to rigorously justify the final step, though. Here is what I have:

Leveraging multiplication by the conjugate, we get: 
  \begin{align*}
\sqrt{n^2 +  n} - n & = \left(\sqrt{n^2 + n} - n\right) \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} + n} \\
& = \frac{n^2 + n + n \sqrt{n^2 + n} - n \sqrt{n^2 + n} - n^2}{\sqrt{n^2 + n} + n} \\
& = \frac{n^2 + n - n^2}{\sqrt{n^2 + n} + n} \\
& = \frac{n}{\sqrt{n^2 + n} + n}
\end{align*}
  Multiplying through by $\frac{\frac{1}{n}}{\frac{1}{n}}$ gives:
  \begin{align*}
& = \frac{\frac{1}{n} \cdot n}{\frac{1}{n} \left(\sqrt{n^2 + n} + n\right)} \\
& = \frac{1}{\frac{1}{n} \sqrt{n^2 + n} + 1} \\
& = \frac{1}{\sqrt{\frac{1}{n^2} \left(n^2 + n\right)} + 1} \\
& = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1}
\end{align*}
  Clearly, a limit of a quotient is the quotient of the limits. The limit of a constant is $1$, so that take scare of the numerator. The limit of a sum is the sum of the limits, so that gets us the limit of $\sqrt{1 + \frac{1}{n}}$ and the limit of $1$, which is a constant. 

However, obvious though it is that this limit is $\frac{1}{2}$ because $\frac{1}{n} \to 0$, I cannot find a theorem in Rudin justifying this. Proving it from scratch would not be too much of a problem, but my question is: is there a general, all-purpose justification for a result like this that says, in effect, that we can "plug in" the limiting argument? If the function were continuous, this would be fine, but we don't have continuity for another chapter.
EDIT: This isn't a duplicate because the linked question does not address my specific questions on this exercise. 
 A: Here's how I solved it back when I self-studied my way through Baby Rudin.  This has never been seen by anyone other than me before, so apologies if there are any errors in it.

As a lemma, we will show that $\lim_{n\rightarrow\infty}\sqrt{n+\alpha}-\sqrt{n}=0$
    for any $\alpha\in\mathbb{P}$. To that end, let $\epsilon\in\mathbb{P}^{\times}$
   be given. Then let $x\in\mathbb{N}$
    be given and assume that $x>\left(\frac{\alpha-\epsilon^{2}}{2\epsilon}\right)^{2}$. As none of the terms that follow can be negative, 
  $$\sqrt{x} > \frac{\alpha-\epsilon^{2}}{2\epsilon}\\
2\epsilon\sqrt{x} > \alpha-\epsilon^{2}\\
\epsilon^{2}+2\epsilon\sqrt{x} > \alpha\\
\epsilon^{2}+2\epsilon\sqrt{x}+x > x+\alpha\\
\sqrt{\epsilon^{2}+2\epsilon\sqrt{x}+x} > \sqrt{x+\alpha}\\
\epsilon+\sqrt{x} > \sqrt{x+\alpha}\\
\epsilon > \sqrt{x+\alpha}-\sqrt{x}$$
   Since $x$
    was an arbitrary “sufficiently large” number, we may conclude that $\lim_{n\rightarrow\infty}\sqrt{n+\alpha}-\sqrt{n}=0$
    as that expression is clearly not strictly negative for any value of $n$.

With that done, the limit in question is quite easily calculated. $$\lim_{n\rightarrow\infty}(\sqrt{n^{2}-n}-n) = 0+\lim_{n\rightarrow\infty}(\sqrt{n^{2}-n}-n)\\
 = \lim_{n\rightarrow\infty}(\sqrt{n^{2}-n+\frac{1}{4}}-\sqrt{n^{2}-n})+\lim_{n\rightarrow\infty}(\sqrt{n^{2}-n}-n)\\
 = \lim_{n\rightarrow\infty}(\sqrt{n^{2}-n+\frac{1}{4}}-\sqrt{n^{2}-n}+\sqrt{n^{2}-n}-n)\\
 = \lim_{n\rightarrow\infty}(\sqrt{n^{2}-n+\frac{1}{4}}-n)\\
 = \lim_{n\rightarrow\infty}((n+\frac{1}{2})-n)=\lim_{n\rightarrow\infty}\frac{1}{2}=\frac{1}{2}$$

So I think that you're in the same boat I was in, where I did a little epsilon wrangling using the definition in 3.1 to justify the part where I wanted to "plug in infinity".
