## Limits by definition

### Task number: 2863

Determine the limits of the following sequences by definition:

#### Variant 1

\(\displaystyle \left\{ \frac1{n} \right\}_{n=1}^\infty\)

#### Resolution

We first conjecture that the limit is \(0\). For any fixed \(\varepsilon>0\) we must find \(n_0\in\mathbb N\) such that \[ \left|\frac{1}{n}-0\right|<\varepsilon \qquad \text{for every}\quad n\in\mathbb N,\ n\geq n_0. \] Choose \(n_0=\left\lfloor\frac{1}{\varepsilon}\right\rfloor+1\). Then for any \(n\in\mathbb N\), \(n\geq n_0\), we have \( n\geq \left\lfloor\frac{1}{\varepsilon}\right\rfloor+1>\frac{1}{\varepsilon}. \) So \( \frac{1}{n}<\varepsilon, \) which we can rewrite in the desired form \( \left|\frac{1}{n}-0\right|<\varepsilon. \)

#### Result

The sequence has limit \(0\).

#### Variant 2

\(\displaystyle \left\{ \frac1{\sqrt{n}} \right\}_{n=1}^\infty\)

#### Resolution

We first guess that the limit is \(0\). So by definition for any \(\varepsilon>0\) we must find \(n_0\in\mathbb N\) such that \[ \left|\frac{1}{\sqrt{n}}-0\right|<\varepsilon \qquad \text{for every}\quad n\in\mathbb N,\ n\geq n_0. \] We can easily rewrite the desired inequality as \(\sqrt{n}>\tfrac{1}{\varepsilon}\) which is equivalent to \(n>\tfrac{1}{\varepsilon^2}\). So we choose \(n_0=\left\lfloor\frac{1}{\varepsilon^2}\right\rfloor+1\). Then for any \(n\in\mathbb N\), \(n\geq n_0\), we have \( n\geq \left\lfloor\frac{1}{\varepsilon^2}\right\rfloor+1>\frac{1}{\varepsilon}. \) So \( \frac{1}{n}<\varepsilon^2, \) which we can rewrite in the desired form \( \left|\frac{1}{\sqrt{n}}-0\right|<\varepsilon. \)

#### Result

The limit of the sequence is 0.

#### Variant 3

\(\displaystyle \left\{ \log n \right\}_{n=1}^\infty\)

#### Resolution

We guess that the limit we seek is \(\infty\). So by definition for any \(K\in\mathbb R\) we must find \(n_0\in\mathbb N\) such that \[ \log{n}>K \qquad \text{for every}\quad n\in\mathbb N,\ n\geq n_0. \] We can easily rewrite the inequality we seek as \(n>e^K\), and so we choose \(n_0=[e^K]+1\). Then for any \(n\in\mathbb N\), \(n\geq n_0\), we have \( n\geq [e^K]+1>e^K, \) and so as desired \( \log{n}>K. \)

#### Result

The sequence is unbounded and has limit \(\infty\).

#### Variant 4

\(\displaystyle \left\{ \frac1{1+n^2} \right\}_{n=1}^\infty\)

#### Hint

We will try to show that the limit is 0.

Clearly \(\frac1{1+n^2}>0\). We choose \(n_0=\left\lceil\frac1{\sqrt\varepsilon}\right\rceil\).

Then \(\frac1{1+n^2}\le \frac1{n_0^2}\le \varepsilon\).

#### Result

The resulting limit is 0.

#### Variant 5

\(\displaystyle \left\{ \frac{n+1}{n+2} \right\}_{n=1}^\infty\)

#### Resolution

We will attempt to show that the limit is 1.

First, we know that \(\frac{n+1}{n+2}<1\). We choose \(n_0=\left\lceil\frac1{\varepsilon}\right\rceil\).

Then \( 1-\frac{n+1}{n+2} =\frac{1}{n+2} <\frac1n \le\frac1{n_0} \le \varepsilon\).

#### Result

The limit is 1.

#### Variant 6

\(\displaystyle \left\{ \sqrt[n]{a}\right\}_{n=1}^\infty\), where \(a\) is a positive real number.

#### Hint

Try raising the sequence to a power. For a simpler estimate it is possible to use Bernoulli's inequality.

#### Resolution

We will first analyze the case \(a\ge1\), since for such an \(a\) we have \(\sqrt[n]{a}\ge1\).

For a given \(\varepsilon\) we choose \(n_0=\left\lceil \frac{a}\varepsilon\right\rceil\). Then for \(n\ge n_0\) we have \(1+n\varepsilon \ge 1+n_0\varepsilon \ge 1+a\).

From there \( a\le 1+n\varepsilon \le (1+\varepsilon)^n\). Taking the root, \(\sqrt[n]{a}\le 1+\varepsilon\) and now \(\sqrt[n]{a}-1\le \varepsilon\).

For \(a<1\) we have \(\sqrt[n]{a}< 1\). Let \(n_0=\left\lceil \frac1{\varepsilon a}\right\rceil\). Now \(a>\frac1{\varepsilon n}\ge \frac1{(1+\varepsilon)^n} \) and so \(\sqrt[n]{a}\ge \frac1{1+\varepsilon}> 1-\varepsilon\). Note: by using the arithmetic of limits we can resolve the second case more easily:

\(\displaystyle \lim_{n\to \infty} \sqrt[n]{a}=\lim_{n\to \infty} \sqrt[n]{\frac1b} = \frac1{\lim_{n\to \infty} \sqrt[n]{b}}=\frac11=1 \) pro \(b=\frac1a>1\).

#### Result

The limit is 1.

#### Variant 7

\(\displaystyle \left\{ \sin \frac1n\right\}_{n=1}^\infty\).

#### Hint

Use an appropriate estimate of the sine function.

#### Resolution

We use the estimate \(\sin x \le x\).

For a given \(\varepsilon\) we choose \(n_0=\left\lfloor\frac{1}{\varepsilon}\right\rfloor+1\). Then for any \(n\in\mathbb N\), \(n\geq n_0\), we have \( \sin \frac1n \le \frac1n \le \frac1{n_0}\le \varepsilon \).

Note: the estimate \(\sin x \le x\) has a clear geometric interpretation. In a right triangle with unit hypotenuse, \(\sin x\) is the length of the opposite side, when \(x\) is measured as the length of an arc with unit radius. From a picture it is evident that the arc must be longer than the side.

#### Result

The limit is 0.