Answer:
The answer is "[tex]\frac{5}{3}[/tex]"
Step-by-step explanation:
[tex]A_n={\frac{13}{5},\frac{18}{8},\frac{23}{11},\frac{28}{14},\frac{33}{17},..................\frac{8+5n}{2+3n}}\\\\[/tex]
[tex]\to A_n=\frac{8+5n}{2+3n}\\\\\to \lim_{n \to \infty} A_n = \lim_{n \to \infty} \ \frac{8+5n}{2+3n}= \lim_{n \to \infty} \ \frac{\frac{8}{n}+5}{\frac{2}{n}+3}=\frac{0+5}{0+3}=\frac{5}{3}[/tex]
In this sequence An is convergents where limits are [tex]\frac{5}{3}[/tex].