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 nāā.
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
