6.5 Illustrations of the Basic Limit Properties.

6.50   Example. In example 6.20, we used the definition of limit to show that

$$\lim\left\{ a^3\left(1+{1\over n}\right)\left(1+{1\over {2n}}\right)\right\}=a^3$$

for all $a \in \mbox{${\mbox{{\bf R}}}^{+}$}$, and claimed that a similar argument shows that

$$\lim\left\{ a^3\left(1-{1\over n}\right)\left(1-{1\over {2n}}\right)\right\}=a^3$$ (6.51)

We will now use the basic properties of limits to prove (6.51). By the product rule and the null sequence rule,

$$\lim\left\{ {1\over {2n}}\right\} = \lim\left\{ {1\over 2}\... ...{1\over 2}\lim\left\{{1\over n}\right\} ={1\over 2}\cdot 0=0.$$

Hence by the sum rule

$$\lim\left\{1-{1\over {2n}}\right\}=\lim\{1\}-\lim\left\{{1\over {2n}}\right\}=1-0=1.$$

By the sum rule and the null sequence rule

$$\lim\left\{1-{1\over n}\right\}=\lim\{1\}-\lim\left\{{1\over n}\right\}=1-0=1.$$

Hence by the product rule,

$$\begin{eqnarray*} \lim \left\{ \left( 1-{1\over n} \right) \cdot \left( 1-{1\ove... ...ft\{ \left( 1-{1\over {2n}} \right) \right\} \\ &=& 1\cdot 1=1. \end{eqnarray*}$$

Now $\{a^3\}$ is a constant sequence, so by the product rule,

$$\begin{eqnarray*} \lim \left\{ a^3\left( 1-{1\over n} \right) \left( 1-{1\over {... ... \left( 1-{1\over {2n}} \right) \right\} \\ &=& a^3\cdot 1=a^3. \end{eqnarray*}$$

6.52   Example. In the previous example, I made at least eight applications of our limit rules. However, the applications are completely mechanical so I will usually not be so careful, and in a situation like this, I will just write

$$\lim\left\{a^3\left(1-{1\over n}\right)\left(1-{1\over {2n}}... ...ight\}=a^3\cdot(1-0)\cdot\left(1-{1\over 2}\cdot 0\right)=a^3.$$ (6.53)

The argument given in equation (6.53) looks remarkably similar to the eighteenth century argument given in example 6.25.

6.54   Example. Let $a$ be a positive number, and let

$$A(a) = \mbox{\rm area}( \{(x,y)\in\mbox{{\bf R}}^2\colon 0\leq x\leq a \mbox{ and } 0\leq y\leq x^2\}).$$

In (2.13), we showed that

$${a^3\over 3}\left(1-{1\over n}\right)\left(1-{1\over {2n}}\r... ...over 3}\left(1+{1\over n}\right) \left(1+{1\over {2n}}\right),$$ (6.55)

for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$, and claimed that these inequalities show that $${A(a) = {a^3\over 3}}$$. Now I want to examine the claim more closely.

In example 6.20 we proved that

$$\lim \left\{ a^3\left( 1+{1\over n}\right) \left(1+{1\over {2n}}\right)\right\} = a^3,$$

and in example 6.50 we proved that

$$\lim\left\{ a^3\left(1-{1\over n}\right)\left(1-{1\over {2n}}\right)\right\}=a^3$$

By applying the squeezing rule to equation 6.55, we see that

$$\lim\{ A(a)\} = a^3,$$

i.e.

$$A(a) = {a^3\over 3}.\mbox{ $\diamondsuit$}$$

6.56   Example. I will now consider the limit

$$\lim\left\{ {{n^2-2n}\over {n^2+3n}}\right\}.$$

Here I cannot apply the quotient rule for sequences, because the limits of the numerator and denominator do not exist. However, I notice that I can simplify my sequence:

$$\left\{ {{n^2-2n}\over {n^2+3n}}\right\}=\left\{ {{n-2}\over {n+3}}\right\}.$$

I will now use a trick. I will factor the highest power of $n$ out of the numerator and denominator:

$$\left\{ {{n-2}\over {n-3}}\right\}=\left\{ {{n\left(1-{2\over... ...}\right\}=\left\{ {{1-{2\over n}}\over {1-{3\over n}}}\right\}.$$

It is now clear what the limit is.

$$\lim\left\{ {{n^2-2n}\over {n^2+3n}}\right\}=\lim\left\{ {{1-... ...er {1-{3\over n}}} \right\}={{1-2\cdot 0}\over {1-3\cdot 0}}=1.$$

6.57   Example. I want to investigate

$$\lim\left\{ {{n-2n^2+3}\over {4+6n+n^2}}.\right\}$$

I'll apply the factoring trick of the previous example.

$$\left\{ {{n-2n^2+3}\over {4+6n+n^2}}\right\} = \left\{ {{n^2\... ...-2+{3\over {n^2}}}\over { {4\over {n^2}}+{6\over n}+1}}\right\}$$

so

$$\begin{eqnarray*} \lim\left\{ {{n-2n^2+3}\over {4+6n+n^2}}\right\} &=& \lim\left... ...\right\}={{0-2+3\cdot 0}\over {4\cdot 0+6\cdot 0+1}} \\ &=& -2. \end{eqnarray*}$$

6.58   Example. I want to find

$$\lim\left\{ {1\over {n+4}}\right\}.$$

I observe that $${ \left\{ {1\over {n+4}} \right\} }$$ is a translate of $${\left\{{1\over n}\right\}}$$ so by the translation rule

$$\lim \left\{ {1\over {n+4}} \right\} = \lim\left\{ {1\over n} \right\}=0.$$

I can also try to do this by my factoring trick:

$$\begin{eqnarray*} \lim\left\{ {1\over {n+4}}\right\} &=& \lim\left\{ {1\over {n\... ...n}}\over {1+{4\over n}}\right\} \\ &=& {0\over {1+4\cdot 0}}=0. \end{eqnarray*}$$

6.59   Exercise. A Find the following limits, or explain why they don't exist.

a)

$${\lim\left\{ 7+{6\over n}+{8\over {\sqrt n}}\right\}}$$

b)

$${\lim\left\{ {{4+{1\over n}}\over {5+{1\over n}}}\right\}}$$

c)

$${\lim\left\{ {{3n^2+n+1}\over {1+3n+4n^2}}\right\}}$$

d)

$${\lim\left\{ {{\left(2+{1\over n}\right)^2+4}\over {\left(2+{1\over n}\right)^3+8}}\right\}}$$

e)

$${\lim\left\{ {{\left(2+{1\over n}\right)^2-4}\over {\left(2+{1\over n}\right)^3-8}}\right\}}$$

f)

$${\lim\left\{ {{8n^3+13n}\over {17+12n^3}}\right\}}$$

g)

$${\lim\left\{ {{8(n+4)^3+13(n+4)}\over {17+12(n+4)^3}}\right\}}$$

h)

$${\lim\left\{ {{n+1}\over {n^2+1}}\right\}}$$.

6.60   Example. Let $a$ be a real number greater than $1$, and let

$$S_a =\{ (x,y)\in\mbox{{\bf R}}^2\colon 1\leq x\leq a \mbox{ and } 0\leq y\leq {1\over {x^2}}\}.$$

In (2.34) we showed that

$${(1 - a^{-1}) \over a^{{1\over n}} }\leq \mbox{\rm area}(S_a... ...(1-a^{-1}) \mbox{ for all } n\in\mbox{${\mbox{{\bf Z}}}^{+}$}.$$ (6.61)

I want to conclude from this that $${\mbox{\rm area}(S_a) =(1-a^{-1})}$$.

By the $n$th root rule, and the quotient and product rules, we have

$$\lim \left\{ {(1 - a^{-1}) \over a^{1\over n} } \right\} = {\... ...\{ a^{1\over n} \} } = { (1 - a^{-1}) \over 1 } = (1 - a^{-1}),$$

and

$$\lim \left\{ a^{1\over n}(1 - a^{-1}) \right \} = \lim \left\... ...t\} \lim\{(1 - a^{-1})\} = 1 \cdot (1 - a^{-1}) = (1 - a^{-1}).$$

By (6.61) and the squeezing rule, we conclude that

$$\lim \{ \mbox{\rm area}(S_a) \} = (1 - a^{-1}),$$

i.e.

$$\mbox{\rm area}(S_a) = (1 - a^{-1}).$$

6.62   Example. Let the sequence $\{a_n\}$ be defined by the rules

$$a_1 = 1,\mbox{{}}$$

$$a_{n+1} = {{a_n^2+2}\over {2a_n}} \mbox{ for } n \geq 1.$$ (6.63)

Thus, for example

$$a_2={{1+2}\over 2}={3\over 2}$$

and

$$a_3={{ {9\over 4}+2}\over 3} ={{17}\over {12}}.$$

It is clear that $a_n>0$ for all $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$. Let $L=\lim\{a_n\}$. Then by the translation rule, $L = \lim\{a_{n+1}\}$ also. From (6.63) we have

$$2a_na_{n+1} = a_n^2+2 \mbox{ for all }n \in \mbox{{\bf Z}}_{\geq 2}.$$

Thus

$$\lim\{2a_na_{n+1}\} = \lim\{ a_n^2+2\},$$

i.e.

$$2\cdot \lim\{a_n\}\lim\{a_{n+1}\} = \lim\{a_n\}^2 + \lim\{2\}.$$

Hence

$$2\cdot L\cdot L = L^2 + 2.$$

Thus $L^2=2$, and it follows that $L=\sqrt 2$ or $L=-\sqrt 2$. But we noticed above that $a_n>0$ for all $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$, and hence by the inequality rule for sequences, $L\geq 0$. Hence we conclude that $L=\sqrt 2$, i.e.,

$$\lim\{a_n\}=\sqrt 2.$$ (6.64)

(Actually there is an error in the reasoning here, which you should try to find, but the conclusion (6.64) is in fact correct. After you have done exercise 6.68, the error should become apparent.)

6.65   Exercise. Use a calculator to find the first six terms of the sequence (6.63). Do all calculations using all the accuracy your calculator allows, and write down the results to all the accuracy you can get. Compare your answers with $\sqrt 2$ (as given by your calculator) and for each term note how many decimal places accuracy you have.

6.66   Example. Let $\{b_n\}$ be the sequence defined by the rules

$$b_1 = 1,$$

$$b_2 = 1,$$

$$b_n = {{1+b_{n-1}}\over {b_{n-2}}} \mbox{ for } n>2.$$ (6.67)

Thus, for example

$$b_3={{1+1}\over 1}=2$$

and

$$b_4={{1+2}\over 1}=3.$$

Notice that $b_n >0$ for all $n$. Let

$$L=\lim\{b_n\}.$$

By the translation rule

$$L=\lim\{b_{n+1}\} \mbox{ and } L=\lim\{b_{n+2}\}.$$

By (6.67) (with $n$ replaced by $n+2$), we have

$$b_n b_{n+2}=1+b_{n+1} \mbox{ for all } n \mbox{ in } \mbox{${\mbox{{\bf Z}}}^{+}$}.$$

Hence

$$\begin{eqnarray*} L^2 &=&\lim\{b_n\}\cdot\lim \{ b_{n+2}\}\\ &=&\lim \left\{ b_{n}b_{n+2} \right\}\\ &=& \lim\{1+b_{n+1}\} = 1+L. \end{eqnarray*}$$

Thus

$$L^2-L-1=0.$$

By the quadratic formula

$$L={{1+\sqrt 5}\over 2} \mbox{ or } L={{1-\sqrt 5}\over 2}.$$

Since $b_n >0$ for all $n$, we have $L\geq 0$, so we have

$$L={{1+\sqrt 5}\over 2}.$$

(This example has the same error as the previous one.)

6.68   Exercise. Repeat exercise 6.65 using the sequence $\{b_n\}$ described in (6.67) in place of the sequence $\{a_n\}$, and $${\left( {{1+\sqrt 5}\over 2}\right)}$$ in place of $\sqrt 2$. After doing this problem, you should be able to point out the error in examples (6.62) and (6.66). (This example is rather surprising. I took it from [14, page 55, exercise 20].)

6.69   Exercise. A For each of the statements below: if the statement is false, give a counterexample; if the statement is true, then justify it by means of limit rules we have discussed.

a)

Let $\{a_n\}$ be a convergent sequence of real numbers. If $a_n>0$ for all $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$, then $\lim\{a_n\}>0$.

b)

Let $\{a_n\}$ and $\{b_n\}$ be real sequences. If $\lim\{a_n\}=0$, then $\lim\{a_n b_n\}=0$.

c)

Let $\{a_n\}$ be a real sequence. If $\lim\{a_n^2\}=1$ then either $\lim\{a_n\}=1$ or $\lim\{a_n\}=-1$.

d)

Let $\{a_n\}$ and $\{b_n\}$ be real sequences. If $\lim\{a_n b_n\}=0$, then either $\lim\{a_n\}=0$ or $\lim\{b_n\}=0$.

6.70   Exercise. Let $a$ and $r$ be positive numbers and let

$$S_0^a[t^r]=\{(x,y)\in\mbox{{\bf R}}^2\colon 0\leq x\leq a \mbox{ and } 0\leq y\leq x^r\}.$$

In (2.4) we showed that

$${{a^{r+1}}\over {n^{r+1}}}\left(1^r+2^r+\cdots +(n-1)^r\right... ..._0^a[t^r])\leq {{a^{r+1}}\over {n^{r+1}}}(1^r+2^r+\cdots +n^r).$$

Use this result, together with the power sums listed in Bernoulli' table to find the area of $S_0^a[t^3]$.

6.71   Theorem ($n$th power theorem.) Let $r$ be a real number such that $\vert r\vert <1$. Then $\lim\{r^n\}=0$.

Proof: Let $L=\lim\{r^{n-1}\}$. Now $\{r^n\}$ is a translate of $\{r^{n-1}\}$, so by the translation theorem

$$\begin{eqnarray*} L&=&\lim\{r^{n-1}\}=\lim\{r^n\}=\lim\{r\cdot r^{n-1}\} \\ &=&\lim\{r\}\lim\{r^{n-1}\}=rL \end{eqnarray*}$$

so we have $L-rL=0$ or

$$L(1-r)=0.$$

We assumed that $\vert r\vert <1$, so $1-r\neq 0$, and hence it follows that $L=0$. $\diamondsuit$

The proof just given is not valid. In fact, the argument shows that $\lim\{r^n\}=0$ whenever $r\neq 1$, and this is certainly wrong when $r=2$. The error comes in the first sentence, `` Let $L=\lim\{r^{n-1}\}$''. The argument works if the sequence $\{r^{n-1}\}$ or $\{r^n\}$ converges. We will now give a second (correct) proof of theorem 6.71.

Second Proof: Let $r$ be a real number with $\vert r\vert <1$. If $r=0$, then $\{r^n\} = \{0\}$ is a constant sequence, and $\lim\{r^n\} = \lim \{0\} = 0$. Hence the theorem holds when $r=0$, and we may assume that $r \neq 0$. Let $\epsilon$ be a generic positive number, If $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ we have

$$(\vert r^n-0\vert<\epsilon)\hspace{1ex}\Longleftrightarrow\hs... ...rrow\hspace{1ex} \left(n\ln(\vert r\vert)<\ln(\epsilon)\right).$$

Now since $\vert r\vert <1$, we know that $\ln(\vert r\vert)<0$ and hence

$$\left( n\ln(\vert r\vert)<\ln(\epsilon)\right)\hspace{1ex}\Lo... ...{1ex}\left(n>{{\ln(\epsilon)}\over {\ln(\vert r\vert)}}\right).$$

By the Archimedean property, there is some positive integer $N(\epsilon)$ such that $${N(\epsilon)>{{\ln(\epsilon)}\over {\ln(\vert r\vert)}}}$$. Then for all $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$

$$\left(n\geq N(\epsilon)\right)\mbox{$\hspace{1ex}\Longrightar... ...e{1ex}\Longrightarrow\hspace{1ex}$}(\vert r^n-0\vert<\epsilon).$$

Hence $\lim\{r^n\}=0$. $\diamondsuit$

6.72   Exercise. Why was it necessary to make $r=0$ a special case in the Second Proof above?