2.5 Limit laws

By now it should be clear that working directly with the definition of a limit is both difficult and time-consuming! For this reason, we wish to develop some theory to do the work for us.

In this section, we introduce and prove various ‘laws’ for manipulating limits which allow us to deduce new limits automatically from known examples. This is achieved using abstract $\varepsilon$ - $N$ arguments like the one used to prove Lemma 2.37.

Addition and multiplication of sequences

The following theorem tells us how convergent sequences behave with respect to the arithmetic operations $+$ and $\,\cdot\,$ .

Theorem 2.38 (Limit laws).

Suppose $(a_{n})_{n\in\mathbb{N}}$ and $(b_{n})_{n\in\mathbb{N}}$ are convergent sequences.

1

The sequence $(a_{n}+b_{n})_{n\in\mathbb{N}}$ is convergent and $\displaystyle\lim_{n\to\infty}(a_{n}+b_{n})=\lim_{n\to\infty}a_{n}+\lim_{n\to% \infty}b_{n}$ ;
2

The sequence $(a_{n}\cdot b_{n})_{n\in\mathbb{N}}$ is convergent and $\displaystyle\lim_{n\to\infty}(a_{n}\cdot b_{n})=\lim_{n\to\infty}a_{n}\cdot% \lim_{n\to\infty}b_{n}$ ;
3

If $b_{n}\neq 0$ for all $n\in\mathbb{N}$ and $\displaystyle\lim_{n\to\infty}b_{n}\neq 0$ , then $(a_{n}/b_{n})_{n\in\mathbb{N}}$ is convergent and

$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\frac{\displaystyle\lim_{n\to\infty}a_{n}% }{\displaystyle\lim_{n\to\infty}b_{n}}.$

Theorem 2.38 provides us with tools which make working with limits a lot simpler.

Example 2.39.

We claim that $\frac{3n^{2}-2n+1}{2n^{2}+n+5}\to\frac{3}{2}$ as $n\to\infty$ .

To see this, we apply the limit laws. For large $n$ , the behaviour is going to be dominated by the terms involving $n^{2}$ . We can highlight this by dividing both the numerator and denominator by $n^{2}$ to give

\frac{3n^{2}-2n+1}{2n^{2}+n+5}=\frac{3-2/n+1/n^{2}}{2+1/n+5/n^{2}}.

Since we know $1/n\to 0$ as $n\to\infty$ , it follows from parts 1 and 2 of Theorem 2.38 that

\lim_{n\to\infty}3-2/n+1/n^{2}=3\qquad\text{and}\qquad\lim_{n\to\infty}2+1/n+5% /n^{2}=2.

Since the limit of the denominator is non-zero, we can apply part 3 of Theorem 2.38 to conclude that

\lim_{n\to\infty}\frac{3n^{2}-2n+1}{2n^{2}+n+5}=\frac{3}{2},

as required.

Exercise 2.40.

Using the limit laws, show the following sequences converge and compute their limits.

(i)

$\displaystyle\Big{(}\frac{10n^{3}+5n+1}{n^{3}+3n^{2}+5}\Big{)}_{n\in\mathbb{N}}$ ;
(ii)

$\displaystyle\Big{(}\frac{2^{n}+3}{2^{n+1}-2}\Big{)}_{n\in\mathbb{N}}$ .

We now turn to discussing the proof of Theorem 2.38. Similar to Lemma 2.37, this involves an abstract $\varepsilon$ - $N$ argument. Our goal is to relate the $\varepsilon$ - $N$ definition of a limit for $(a_{n})_{n\in\mathbb{N}}$ and $(b_{n})_{n\in\mathbb{N}}$ to the $\varepsilon$ - $N$ definition applied to the new sequences $(a_{n}+b_{n})_{n\in\mathbb{N}}$ , $(a_{n}\cdot b_{n})_{n\in\mathbb{N}}$ and $(a_{n}/b_{n})_{n\in\mathbb{N}}$ .

Before presenting the details of the proof of Theorem 2.38, we briefly sketch the key idea for part 2 (which is the hardest part). The proof hinges on the elementary identity

(2.4) (2.4)

a_{n}\cdot b_{n}-a\cdot b=b(a_{n}-a)+a_{n}(b_{n}-b)=b(a_{n}-a)+(a_{n}-a)(b_{n}% -b)+a(b_{n}-b).

This allows us to relate the difference $|a_{n}\cdot b_{n}-a\cdot b|$ appearing in the $\varepsilon$ - $N$ definition for $a_{n}\cdot b_{n}\to a\cdot b$ to the differences $|a_{n}-a|$ and $|b_{n}-b|$ appearing in the $\varepsilon$ - $N$ definition for $a_{n}\to a$ and $b_{n}\to b$ . Similar formulæ are also used to prove part 1 and part 3.

Proof (of Theorem 2.38).

Let $\displaystyle a:=\lim_{n\to\infty}a_{n}$ and $\displaystyle b:=\lim_{n\to\infty}b_{n}$ .

1.

Let $\varepsilon>0$ be given. Since $a_{n}\to a$ and $b_{n}\to b$ as $n\to\infty$ , by the $\varepsilon$ - $N$ definition of a limit, there exists some $N_{1}$ , $N_{2}\in\mathbb{N}$ such that

$|a_{n}-a|<\frac{\varepsilon}{2}\quad\text{for all $n>N_{1}$}\qquad\text{and}% \qquad|b_{n}-b|<\frac{\varepsilon}{2}\quad\text{for all $n>N_{2}$.}$

Choose $N:=\max\{N_{1},N_{2}\}$ . By the triangle inequality,

$|a_{n}+b_{n}-(a+b)|\leq|a_{n}-a|+|b_{n}-b|<\frac{\varepsilon}{2}+\frac{% \varepsilon}{2}=\varepsilon\qquad\text{for all $n\in\mathbb{N}$ with $n>N$.}$

Thus, by the $\varepsilon$ - $N$ definition of a limit, $a_{n}+b_{n}\to a+b$ as $n\to\infty$ .

By taking absolute values of both sides of (2.4) and applying the triangle inequality, we have

(2.5) (2.5)

|a_{n}\cdot b_{n}-a\cdot b|\leq|b||a_{n}-a|+|a_{n}-a||b_{n}-b|+|a||b_{n}-b|

Let $\varepsilon>0$ be given. We will further assume $0<\varepsilon\leq 1$ . Our goal is to bound the right-hand side of (2.5) by $\varepsilon$ , which we can do by bounding the $|a_{n}-a|$ and $|b_{n}-b|$ terms appropriately. Since $a_{n}\to a$ and $b_{n}\to b$ as $n\to\infty$ , by the $\varepsilon$ - $N$ definition of a limit, there exists some $N_{1}$ , $N_{2}\in\mathbb{N}$ such that

|a_{n}-a|<\frac{\varepsilon}{3(|b|+1)}\quad\text{for all $n>N_{1}$}\quad\text{% and}\quad|b_{n}-b|<\frac{\varepsilon}{3(|a|+1)}\quad\text{for all $n>N_{2}$.}

Choose $N:=\max\{N_{1},N_{2}\}$ . Recalling (2.5), for all $n\in\mathbb{N}$ with $n>N$ it follows that

$\displaystyle\|a_{n}\cdot b_{n}-a\cdot b\|$	$\displaystyle\leq\|b\|\|a_{n}-a\|+\|a_{n}-a\|\|b_{n}-b\|+\|a\|\|b_{n}-b\|$
	$\displaystyle<\|b\|\cdot\frac{\varepsilon}{3(\|b\|+1)}+\frac{\varepsilon}{3(\|b\|+1)% }\cdot\frac{\varepsilon}{3(\|a\|+1)}+\|a\|\cdot\frac{\varepsilon}{3(\|a\|+1)}$
	$\displaystyle<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3% }=\varepsilon.$

We have shown that the $\varepsilon$ - $N$ definition holds for all $0<\varepsilon\leq 1$ . However, by applying the above to $\min\{\varepsilon,1\}$ , this implies that the definition holds for all $\varepsilon>0$ . Thus, $a_{n}\cdot b_{n}\to a\cdot b$ as $n\to\infty$ .

3.

We first show that the sequence $(1/b_{n})_{n\in\mathbb{N}}$ is convergent and $1/b_{n}\to 1/b$ as $n\to\infty$ . The idea behind the proof is similar to that used above; we leave the details to Exercise 2.41 below.

Since $a_{n}\to a$ as $n\to\infty$ and $1/b_{n}\to 1/b$ as $n\to\infty$ , we can use the result from part 2 to conclude that $a_{n}/b_{n}\to a/b$ as $n\to\infty$ , as required.

∎

Exercise 2.41.

Let $(b_{n})_{n\in\mathbb{N}}$ be a convergent sequence with $b_{n}\neq 0$ for all $n\in\mathbb{N}$ and suppose $b:=\lim_{n\to\infty}b_{n}\neq 0$ . Show that $1/b_{n}\to 1/b$ as $n\to\infty$ , and thereby complete the proof of part 3 of Theorem 2.38.

The squeeze theorem

Theorem 2.38 describe how limits interact with the arithmetic operations $+$ and $\,\cdot\,$ . We can also consider how limits interact with the order $\leq$ on $\mathbb{R}$ .

Theorem 2.42 (Squeeze theorem).

Suppose $(a_{n})_{n\in\mathbb{N}}$ , $(\ell_{n})_{n\in\mathbb{N}}$ , $(u_{n})_{n\in\mathbb{N}}$ are three sequences such that:

(i)

$\ell_{n}\leq a_{n}\leq u_{n}$ for all $n\in\mathbb{N}$ , and
(ii)

$(\ell_{n})_{n\in\mathbb{N}}$ and $(u_{n})_{n\in\mathbb{N}}$ are convergent and $\displaystyle\lim_{n\to\infty}\ell_{n}=\lim_{n\to\infty}u_{n}=a$ for some $a\in\mathbb{R}$ .

Then $(a_{n})_{n\in\mathbb{N}}$ is convergent and $\displaystyle\lim_{n\to\infty}a_{n}=a$ .

Example 2.43.

Consider the sequence $\big{(}\frac{\sin n}{n}\big{)}_{n\in\mathbb{N}}$ .

The terms of the sequence satisfy $-\frac{1}{n}\leq\frac{\sin n}{n}\leq\frac{1}{n}$ for all $n\in\mathbb{N}$ , since we know that $-1\leq\sin x\leq 1$ for all $x\in\mathbb{R}$ .⁵⁵ 5 Here (and also in the worksheets) we use the $\sin$ function, which we haven’t formally defined (but which, of course, you are familiar with from high school). Since we’ve gone to great lengths to formally define other expressions such as $\sqrt{n}$ , it seems remiss not to do the same for $\sin$ . However, for now, we shall assume a working knowledge of trigonometric functions, in order to furnish our theory with additional examples. We shall come back to the question of their formal definition later. We know from Example 2.25 that $1/n\to 0$ as $n\to\infty$ , and it follows from the limit laws that $-1/n\to 0$ as $n\to\infty$ .

Hence, by the squeeze theorem (applied to $a_{n}=\frac{\sin n}{n}$ , $\ell_{n}=-\frac{1}{n}$ and $u_{n}=\frac{1}{n}$ ), we conclude that $\frac{\sin n}{n}\to 0$ as $n\to\infty$ .

Proof (of Theorem 2.42).

Let $\varepsilon>0$ so that, by the $\varepsilon$ - $N$ definition of a limit, there exist $N_{1}$ , $N_{2}\in\mathbb{N}$ such that

|\ell_{n}-a|<\varepsilon\quad\text{for all $n>N_{1}$}\qquad\text{and}\qquad|u_% {n}-a|<\varepsilon\quad\text{for all $n>N_{2}$.}

Let $N:=\max\{N_{1},N_{2}\}$ . If $n\in\mathbb{N}$ with $n>N$ , then it follows that

a-\varepsilon<\ell_{n}\leq a_{n}\leq u_{n}<a+\varepsilon.

In particular, $|a_{n}-a|<\varepsilon$ . Thus, by the $\varepsilon$ - $N$ definition of a limit, $a_{n}\to a$ as $n\to\infty$ . ∎

Exercise 2.44.

Use the squeeze theorem to show $\frac{(-1)^{n}}{2^{n}+5}\to 0$ as $n\to\infty$ .

Exercise 2.45.

Let $A\subseteq\mathbb{R}$ be a nonempty set which is bounded above, so that $s:=\sup A$ exists by the completeness axiom.

(i)

For each $n\in\mathbb{N}$ , show there exists some $a_{n}\in A$ such that $s-1/n<a_{n}\leq s$ .
(ii)

Conclude that the sequence $(a_{n})_{n\in\mathbb{N}}$ satisfies $a_{n}\in A$ for all $n\in\mathbb{N}$ and $a_{n}\to s$ as $n\to\infty$ .

Exercise 2.46.

Let $(a_{n})_{n\in\mathbb{N}}$ and $(b_{n})_{n\in\mathbb{N}}$ be convergent sequences.

(i)

Suppose $a_{n}\leq b_{n}$ for all $n\in\mathbb{N}$ . Show that $\displaystyle\lim_{n\to\infty}a_{n}\leq\lim_{n\to\infty}b_{n}$ .
(ii)

Suppose $a_{n}<b_{n}$ for all $n\in\mathbb{N}$ . Is it necessarily true that $\displaystyle\lim_{n\to\infty}a_{n}<\lim_{n\to\infty}b_{n}$ ?

$\displaystyle\|a_{n}\cdot b_{n}-a\cdot b\|$	$\displaystyle\leq\|b\|\|a_{n}-a\|+\|a_{n}-a\|\|b_{n}-b\|+\|a\|\|b_{n}-b\|$
	$\displaystyle<\|b\|\cdot\frac{\varepsilon}{3(\|b\|+1)}+\frac{\varepsilon}{3(\|b\|+1)% }\cdot\frac{\varepsilon}{3(\|a\|+1)}+\|a\|\cdot\frac{\varepsilon}{3(\|a\|+1)}$
	$\displaystyle<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3% }=\varepsilon.$