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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 $\displaystyle \lim_{n \rightarrow \infty}x_n=c$?

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    $\begingroup$ You want prove using the definition of limit? $\endgroup$ – Juniven Apr 1 '17 at 13:52
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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 $$\lim_{n\to\infty}x_n=c.$$

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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.

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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 $$\lim_{n\to\infty}f(n) = c$$

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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?

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