Answer
(a) $M=999$
$\mathop {\lim }\limits_{n \to \infty } {a_n} = 1$
(b) $M=99999$
$\mathop {\lim }\limits_{n \to \infty } {a_n} = 1$
Work Step by Step
(a) The limit definition requires us to find, for $\varepsilon = 0.001 > 0$, a number $M$ such that
(1) ${ \ \ \ \ \ }$ $\left| {{a_n} - 1} \right| \le 0.001$ for all $n \ge M$.
We have
$\left| {{a_n} - 1} \right| = \left| {\frac{n}{{n + 1}} - 1} \right| = \left| {\frac{{n - n - 1}}{{n + 1}}} \right| = \frac{1}{{n + 1}}$
Therefore, $\left| {{a_n} - 1} \right| \le 0.001$ if
$\frac{1}{{n + 1}} \le 0.001$,
$\frac{1}{{0.001}} \le n + 1$
or $n \ge 999$
Thus, equation (1) is valid with $M=999$. By the limit definition, this proves that $\mathop {\lim }\limits_{n \to \infty } {a_n} = 1$.
(b) Using the same method as in part (a):
We have
$\left| {{a_n} - 1} \right| = \left| {\frac{n}{{n + 1}} - 1} \right| = \left| {\frac{{n - n - 1}}{{n + 1}}} \right| = \frac{1}{{n + 1}}$
Therefore, $\left| {{a_n} - 1} \right| \le 0.00001$ if
$\frac{1}{{n + 1}} \le 0.00001$,
$\frac{1}{{0.00001}} \le n + 1$,
or $n \ge 99999$
Thus, equation (1) is valid with $M=99999$. By the limit definition, this proves that $\mathop {\lim }\limits_{n \to \infty } {a_n} = 1$.
