Answer
By limit definition: $\mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + {n^{ - 1}}}} = 1$
Work Step by Step
The limit definition requires us to find, for every $\varepsilon > 0$, a number $M$ such that
(1) ${\ \ \ \ \ }$ $\left| {\frac{n}{{n + {n^{ - 1}}}} - 1} \right| < \varepsilon $ ${\ \ }$ for all $n>M$.
We have
$\left| {\frac{{n - n - {n^{ - 1}}}}{{n + {n^{ - 1}}}}} \right| < \varepsilon $,
$\left| {\frac{{ - {n^{ - 1}}}}{{n + {n^{ - 1}}}}} \right| < \varepsilon $,
$\left| {\frac{{ - 1}}{{{n^2} + 1}}} \right| < \varepsilon $,
$\frac{1}{{{n^2} + 1}} < \varepsilon $,
${n^2} > \frac{1}{\varepsilon } - 1$,
$n > \sqrt {\frac{1}{\varepsilon } - 1} $.
We can choose $\varepsilon > 0$ and $M = \sqrt {\frac{1}{\varepsilon } - 1} $ so that equation (1) is valid. By limit definition, this proves that $\mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + {n^{ - 1}}}} = 1$.
