$\forall n\in\mathbb{N}$, $a_n\not=0$, $\lim_{n\to\infty} a_n=0$. Find $\lim_{n\to\infty} \frac{\sqrt{3+a_n}-\sqrt{3}}{a_n}$ I already multiplied by the conjugate so I get
$$\lim_{n\to\infty} \frac{1}{\sqrt{3+a_n}+\sqrt{3}}$$
but I dont know how to prove that $\lim_{n\to\infty} \sqrt{3+a_n}=\sqrt{3}$.
Also I already tried with the definition with the limit $\frac{1}{2\sqrt{3}}$, but is more hard. Can anyone help me? Also important to say that $a_n$ is a sequence.
We know that for positive values, $\sqrt{ \cdot }$ is a continuous function, meaning that the limit of the function is the function of the limit.  So $\lim_{n \rightarrow \infty} \sqrt{3+a_n}=\sqrt{\lim_{n \rightarrow \infty}(3+a_n)}$.  Find $\lim_{n \rightarrow \infty} 3+a_n$ and you should be good!
$$\begin {align} \lim_{n\to \infty}\sqrt {3+a_n}=\sqrt {\lim_{n\to \infty}\left(3+a_n\right)}=\sqrt {\lim_{n\to \infty }3+\lim_{n\to \infty} a_n}=\sqrt {3+0}=\sqrt 3\end{align} $$