Answer
(a) Since the sequence is bounded below by $7$ and decreasing, its limit exists by theorem 9.5
(b) The graph appears below, the limit is 7
Work Step by Step
(a) We show that $a_{n}$ is bounded below by $7$:
We know that for all $n \geq 1$ , $\frac{1}{n} \geq 1 $ because $1$ divided by a positive number is a positive number. Thus, since $\frac{1}{n} \geq 1 $, by adding $7$ to both sides:
$a_{n} =$ $7 + \frac{1}{n} \geq 7 +1
= 8 > 7$, so $a_{n}$ is bounded below by $7$.
Finally, we know that $a_{n}$ is decreasing because for all $n$:
$n+1 \geq n \implies $ (by dividing) $\frac{(n+1)}{n} \geq 1 \implies 1/n \geq 1/(n+1) \implies 7 +1/n \geq 7+ 1/(n+1) \implies $
$a_{n} \geq a_{n+1}$
Since it is decreasing and bounded below, it has a limit.
(b) The desired graph appears above. Since the sequence decreases towards seven as $n$ increases, the limit is $7$
