Show that any null sequence $(a_n)_{n\in\mathbb R}, \ a_n\neq 0$ fulfils
$$\lim_{n\to\infty}\frac{\sqrt{1+a_n}-1}{a_n}=\frac{1}{2}\tag{1}$$
Was wondering if my approach is valid.
Let's assume that $\lim_{n\to\infty}a_n$ exists and that $\lim_{n\to\infty}=0$ then we know that $\lim_{n\to\infty} (\sqrt{1+a_n} -1)$ also exists. So we can write:
$$\lim_{n\to\infty}\frac{\sqrt{1+a_n}-1}{a_n}=\frac{\lim_{n\to\infty}\sqrt{1+a_n}-1}{\lim_{n\to\infty}a_n}=\frac{1}{2}\tag{2}$$
Now we know that we can rewrite this to
$$\lim_{n\to\infty}\sqrt{1+a_n}-1 = \frac{1}{2}\lim_{n\to\infty}a_n\tag{3}$$
We use the fact that the square root is continuous:
$$\lim_{n\to\infty}\sqrt{1+a_n}-1=\sqrt{\lim_{n\to\infty}1+a_n}-1 = \frac{1}{2}\lim_{n\to\infty}a_n\tag{4}$$
We get
$$\sqrt{1+0}-1=\frac{1}{2}\cdot 0\tag{5}$$
So the equation does hold for any null sequence $a_n$.
Question: Is there any flaw here?
Regarding your question whether there is a flaw: You need to be careful with mixing the assumptions and what you actually want to prove. You cannot use $$\lim_{n\to\infty}\frac{\sqrt{1+a_n}-1}{a_n}=\frac{1}{2}$$ in (2) because this is what you actually want to prove! Even if the following steps are equivalences (and you have to be very careful there, in particular with square roots), it is confusing at best and therefore not a very solid proof. Therefore my advice: Try writing proofs in a way that you start with the assumptions and end up with the result instead of mixing the result in somewhere in the middle and conclude with a tautology. This will save you from a lot of trouble and also makes your proofs much easier to read.