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# 6.5 Illustrations of the Basic Limit Properties.

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

for all , and claimed that a similar argument shows that
 (6.51)

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

Hence by the sum rule

By the sum rule and the null sequence rule

Hence by the product rule,

Now is a constant sequence, so by the product rule,

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
 (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 be a positive number, and let

In (2.13), we showed that
 (6.55)

for all , and claimed that these inequalities show that . Now I want to examine the claim more closely.

In example 6.20 we proved that

and in example 6.50 we proved that

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

i.e.

6.56   Example. I will now consider the limit

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:

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

It is now clear what the limit is.

6.57   Example. I want to investigate

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

so

6.58   Example. I want to find

I observe that is a translate of so by the translation rule

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

6.59   Exercise. A Find the following limits, or explain why they don't exist.
a)
b)
c)
d)
e)
f)
g)
h)
.

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

In (2.34) we showed that
 (6.61)

I want to conclude from this that .

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

and

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

i.e.

6.62   Example. Let the sequence be defined by the rules
 (6.63)

Thus, for example

and

It is clear that for all in . Let . Then by the translation rule, also. From (6.63) we have

Thus

i.e.

Hence

Thus , and it follows that or . But we noticed above that for all in , and hence by the inequality rule for sequences, . Hence we conclude that , i.e.,
 (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 (as given by your calculator) and for each term note how many decimal places accuracy you have.

6.66   Example. Let be the sequence defined by the rules
 (6.67)

Thus, for example

and

Notice that for all . Let

By the translation rule

By (6.67) (with replaced by ), we have

Hence

Thus

Since for all , we have , so we have

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

6.68   Exercise. Repeat exercise 6.65 using the sequence described in (6.67) in place of the sequence , and in place of . 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 be a convergent sequence of real numbers. If for all in , then .
b)
Let and be real sequences. If , then .
c)
Let be a real sequence. If then either or .
d)
Let and be real sequences. If , then either or .

6.70   Exercise. Let and be positive numbers and let

In (2.4) we showed that

Use this result, together with the power sums listed in Bernoulli' table to find the area of .

6.71   Theorem (th power theorem.) Let be a real number such that . Then .

Proof: Let . Now is a translate of , so by the translation theorem

so we have or

We assumed that , so , and hence it follows that .

The proof just given is not valid. In fact, the argument shows that whenever , and this is certainly wrong when . The error comes in the first sentence,  Let ''. The argument works if the sequence or converges. We will now give a second (correct) proof of theorem 6.71.

Second Proof: Let be a real number with . If , then is a constant sequence, and . Hence the theorem holds when , and we may assume that . Let be a generic positive number, If we have

Now since , we know that and hence

By the Archimedean property, there is some positive integer such that . Then for all in

Hence .

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

Next: 6.6 Geometric Series Up: 6. Limits of Sequences Previous: 6.4 Properties of Limits.   Index
Ray Mayer 2007-09-07