Calculate the limit of a real sequence

I need to show that the following real sequence is convergent. Let $r,l>0$ be constant then the sequence $(a_n)_{n\in \mathbb{N}}$ is defined by
$$a_n=sin^{-1}\left(\frac{r}{2n}\right)\sqrt{1+n^2l^2}.$$

Furthermore, I also need to determine the limit (which is $\frac{r}{2}l$).

Thank you in advance for your answers and ideas.

Answer

Since $\lim _{ n\rightarrow 0 }{ \frac { \arcsin { n } }{ n } } =1,\lim _{ n\rightarrow \infty }{ \frac { \arcsin { \left( \frac { r }{ 2n } \right) } }{ \left( \frac { r }{ 2n } \right) } } =1$ so

$$\lim _{ n\rightarrow \infty }{ \arcsin { \left( \frac { r }{ 2n } \right) } \sqrt { 1+n^{ 2 }l^{ 2 } } } =\lim _{ n\rightarrow \infty }{ \frac { \arcsin { \left( \frac { r }{ 2n } \right) } }{ \frac { r }{ 2n } } \cdot \frac { r }{ 2n } \cdot \sqrt { 1+n^{ 2 }l^{ 2 } } } =\\ =\lim _{ n\rightarrow \infty }{ \frac { r }{ 2n } \cdot n\cdot \sqrt { \frac { 1 }{ { n }^{ 2 } } +l^{ 2 } } } =\color{red}{\frac { lr }{ 2 }}$$