Determining the limit: What does it mean when one obtains zero in the calculation? Consider the sequence $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\\$. To determine the limit I did the following:
\begin{aligned}
a_{n} &=\left(\sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\right) \frac{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}} \\[10pt]
&=\frac{2 \sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}=\frac{2 \sqrt{n}}{\sqrt{n} \sqrt{1+\frac{1}{\sqrt{n}}}+\sqrt{n} \sqrt{1-\frac{1}{\sqrt{n}}}} \\[10pt]
&=\frac{2}{\sqrt{1+\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}+\sqrt{1-\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}} \\[10pt] &= \dfrac{2}{2} = 1.
\end{aligned}
However, my first thought was $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}} = \sqrt{n}\left(\sqrt{1 + \dfrac{1}{\sqrt{n}}}\space - \sqrt{1 - \dfrac{1}{\sqrt{n}}}\right) = \sqrt{n}\space (1-1) = 0$
with the same argument as above. Is it correct that one can't make a statement about the convergence in the latter calculation because we have $\infty \cdot 0$ ? If yes, why exactly is this the case?
Edit: To see that $\lim\limits_{n\to \infty} \dfrac{1}{\sqrt{n}} = 0$ pick some arbitrary $\epsilon > 0$. We want to find a $N$ s.t. $\forall n\geq N\colon |\dfrac{1}{\sqrt{n}} - 0|< \epsilon \Longleftrightarrow \dfrac{1}{\sqrt{n}} < \epsilon \Longleftrightarrow n >\dfrac{1}{\epsilon^2}$ meaning we can choose $N = \dfrac{1}{\epsilon^2} + 1$ for example.

When one factor of a product goes to $\infty$ and the other goes to 0, you can't immediately conclude what the limit is. The easiest way to see this is to consider $\lim_{n \to \infty} n \cdot \frac{1}{n}$.  Now replace the 1 by any other number.
The error in your calculation is that when you evaluate a limit by substitution, you have to substitute all instances of $n$ at the same time.  In your calculation, in the second to last = sign, you substitute some instances to get $1-1$ while leaving the $\sqrt{n}$ unsubstituted.  That is wrong because when you take a limit, all instances of $n$ are tending to the limit at the same time.
