IDNLearn.com: Your go-to resource for finding expert answers. Ask your questions and get detailed, reliable answers from our community of knowledgeable experts.
Sagot :
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].
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.