$$\underset{n\to \infty }{lim} \frac{n+3}{n^2+3n+2}=$$

$$\underset{n\to \infty }{lim} \frac{n·\left(1+{\displaystyle \frac{3}{n}}\right)}{n^2\left(1+{\displaystyle \frac{3n}{n^2}}+{\displaystyle \frac{2}{n^2}}\right)}=$$

$$\underset{n\to \infty }{lim} \frac{\left(1+{\displaystyle \frac{3}{n}}\right)}{n\left(1+{\displaystyle \frac{3}{n}}+{\displaystyle \frac{2}{n^2}}\right)}=$$

$$=\frac{\left(1+{\displaystyle \frac{3}{\infty }}\right)}{\infty ·\left(1+{\displaystyle \frac{3}{\infty }}+{\displaystyle \frac{2}{{\infty }^2}}\right)}$$

$$=\frac{\left(1+0\right)}{\infty ·\left(1+{\displaystyle 0}+{\displaystyle 0}\right)}$$

$$=\frac{1}{\infty ·1}$$

$$=\frac{1}{\infty }$$