Convergence of root sequence. I need help to make my proof absolutely correct. 
We got $a_n = \sqrt{n} \cdot (\sqrt{n+3}-\sqrt{n})$. Proof that the sequence diverges. 
Proof: 
$\lim\limits_{n \rightarrow \infty} \sqrt{n} = \infty $
$ \sqrt{n+3} > \sqrt{n}  \Rightarrow (\sqrt{n+3} - \sqrt{n}) > 0 \Rightarrow \lim\limits_{n \rightarrow \infty} (\sqrt{n+3} - \sqrt{n}) = \infty $
$\Rightarrow \lim\limits_{n \rightarrow \infty} \sqrt{n} \cdot (\sqrt{n+3}-\sqrt{n}) = \infty $
I do not know if my proof is perfectly correct. Please do not tell me that there are complete different ways to solve this because i am not allowed to use higher technics in my exam. So please help me to improve my solution.
 A: Hint
$$\sqrt{n}\cdot (\sqrt{n+3}-\sqrt{n})=\frac{\sqrt{n}\cdot(\sqrt{n+3}-\sqrt{n})(\sqrt{n+3}+\sqrt{n})}{(\sqrt{n+3}+\sqrt{n})}=\frac{3\sqrt{n}}{(\sqrt{n+3}+\sqrt{n})}=\frac{3}{\sqrt{1+\frac{3}{n}}+1}$$
Fot the last equality just divide both terms by $\sqrt{n}$. 
Now you can do $n \rightarrow \infty$. 
Can you finish?
A: Please note that while it is true that
$$\sqrt{n+3} - \sqrt{n} > 0$$
it is false that 
$$\lim_{x\to \infty} \sqrt{n+3} - \sqrt{n} = \infty \tag{1}$$
given that we actually have
$$\lim_{x\to \infty} \sqrt{n+3} - \sqrt{n} = 0 \tag{2}$$
So the limit you want to find is actually a limit of the form $0 \cdot \infty$ because $\sqrt{n} \to \infty$ when $n \to \infty$.
Therefore we must do something else to find the limit. A suggested approach would be to multiply and divide by $\sqrt{n+3} + \sqrt{n}$ giving
$$\lim_{n \to \infty} \sqrt{n}\cdot \frac{(\sqrt{n+3}-\sqrt{n})(\sqrt{n+3}+\sqrt{n})}{\sqrt{n+3}+\sqrt{n}}$$
You should be able to simplify the numerator to get
$$\lim_{n \to \infty} \frac{3\sqrt{n}}{\sqrt{n+3}+\sqrt{n}}$$
Can you solve this new limit? First, convince yourself that when $n$ goes to infinity, $\sqrt{n+3} \approx \sqrt{n}$
