How to prove that the limit of sequence $\langle x_n \rangle$ such that $x_n=c+\dfrac{1}{nr}$ is $c$? Let $c \in \mathbb R$.
Let $\langle x_n \rangle$ be a sequence of real numbers such that $\forall n \in \mathbb N^*:x_n=c+\dfrac{1}{nr}$, where $r$ is some constant real number.
How to prove that $\displaystyle \lim_{n \rightarrow \infty}x_n=c$?
 A: Hint: Do you know $\lim_{n\to\infty} \frac{1}{n}$? And how to calculate the sums and products of known limits? Once you know those things you can piece together your answer.
A: By first principles:
Let $\epsilon\gt0$ be given.
The limit is $c$ if $\exists M\in \mathbb R^+$ such that
$$|f(n)-c|\lt \epsilon$$ whenever $n\gt M$.
Here,
$$|f(n)-c|=\left|c+\frac{1}{nr}-c\right|=\left|\frac{1}{nr}\right|$$
By the Archimedean property, $\exists n_0\in \mathbb N$ such that
$$n_0|r|\gt\frac{1}{\epsilon}$$
Thus,
$$\left|\frac{1}{n_0r}\right|\lt \epsilon$$
For all $n\gt n_0$, $$\left|\frac{1}{nr}\right|\lt\left|\frac{1}{n_0r}\right|\lt\epsilon$$
Thus, $n_0$ satisfies the criteria for $M$, and it can be concluded that
$$\lim_{n\to\infty}f(n) = c$$
A: We say $L$ is the limit of $a_n$ if for any $\epsilon > 0$ exists $N_0$ such that for all $n>N_0$,  $|a_n - L| < \epsilon$
$|a_n-L|=|c+\frac{1}{nr}-c|=|\frac{1}{nr}|<\epsilon$
Can you continue this?
A: Let $\epsilon>0$. By using the Archimedean Property, there exists $N\in \Bbb N$ such that $\frac{1}{N}<|r|\epsilon$. Thus, if $n\geq N$ then
$$
\begin{align}
\big|x_n-c\big|&=\bigg|c-\frac{1}{nr}-c\bigg|\\
&=\frac{1}{|r|}\cdot\frac{1}{n}\\
&\leq \frac{1}{|r|}\cdot\frac{1}{N}\\
&< \frac{1}{|r|}\cdot|r|\epsilon\\
&=\epsilon.
\end{align}$$
Hence, using the definition of limit, we get
$$\lim_{n\to\infty}x_n=c.$$
