Given the sequence,
[tex](\frac{5n-1}{n+1})[/tex]
We can find the solution to the question below.
Explanation
1) The sequence converges.
2) This is because the limit of the sequence exists as n→∞.
We can find the limit below.
[tex]\begin{gathered} \lim_{n\to\infty\:}\left(\frac{5n-1}{n+1}\right) \\ divide\text{ the numerator and denominator by n} \\ \lim_{n\to\infty\:}\left(\frac{5-\frac{1}{n}}{1+\frac{1}{n}}\right) \\ Recall;\lim_{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim_{x\to a}f\left(x\right)}{\lim_{x\to a}g\left(x\right)},\:\quad\lim_{x\to a}g\left(x\right)0 \\ \frac{\lim_{n\to\infty\:}\left(5-\frac{1}{x}\right)}{\lim_{n\to\infty\:}\left(1+\frac{1}{x}\right)}=\frac{5}{1}=5 \end{gathered}[/tex]
Answer: The limit of the sequence is 5
