Limits

Section 2.1 Problems

2.2 Sequences

2.2.2 Limits

in which the value ofa_n+1depends on the valuesa_n anda_n−1—that is, on the values one and two time steps back. To determine the values of successive members of a sequence given in recursive form, we need to specify an initial valuea₀if we start the sequence atn=^{0 (or}a₁if we start the sequence atn=^1).

In the notation of this section, the exponential growth of the previous section is given explicitly by

an =a₀Rⁿ, n=⁰,¹,², . . . and recursively by

a_n+1 = Ran, n=⁰,¹,², . . .

Note that, in the recursive definition, the initial valuea0needs to be specified.

Solution

Successive terms ofa_n, namely,

1,2,4,8,16,32, . . .

indicate that the terms continue to grow. Hence,a_n goes to infinity asn → ∞, and we can write lim_n→∞a_n = ∞. Since infinity (∞) is not a real number, we say that the limit does not exist.

Let’s look at one more example of a limit that exists before we give a formal definition.

EXAMPLE 8 Find

n→∞lim n+¹

n

Solution

Starting withn=1 and computing successive terms, we find that

2,³ 2,⁴

3,⁵ 4,⁶

5, . . .

We see that the terms get closer and closer to 1, and, indeed,

n→∞lim n+¹

n =¹

The way we solved the first four examples is unsatisfying: We guessed the limiting values. How do we know that our guesses are correct? There is a formal definition of limits that can be used to compute them. However, except in the simplest cases, the formal definition is quite cumbersome to use. Fortunately, there are mathematical laws that build on simple limits (which can be computed from the formal definition).

We will first discuss the formal definition (as an optional topic) and then introduce the limit laws.

Formal Definition of Limits (Optional) Example 8 will motivate the formal definition of limits. When we guessed the limit in Example 8, we realized that successive terms approached 1. This means that no matter how small an interval about 1 we choose, all points must lie in this interval for all sufficiently large values ofn. Graphically, the points of the graph ofa_nmust lie between the two dashed lines in Figure 2.10 for all large enough values ofn, no matter how close those lines are to the horizontal line at height 1.

1.4 1.2 1 0.8

0.4 0.6

0 an

n

5 10 15 20

Figure 2.10 Convergence of the sequencean= ⁿ⁺_n¹ toa=1.

Translating this condition into a formal statement for the general case, we arrive at the following definition:

Formal Definition of Limits The sequence {an} ^{has limit} a, written as lim_n→∞an =a, if, for every >0, there exists an integer N such that

|an−a|< ^whenevern> N

If the limit exists, the sequence is called convergent and we say that an

convergestoaas ntends to infinity. If the sequence has no limit, it is called divergent.

The value ofN will typically depend on : The smaller is, the larger N is. We illustrate the concept of a converging sequence in Figure 2.11. The horizontal dashed lines are at heightsa+ ^anda− , respectively. They form a strip of width 2 centered at the horizontal line at height a. Points an within this strip satisfy the inequality

|an−a|< . For a sequence to be convergent, we require thatallpointsanlie in this strip forallnsufficiently large (namely, larger than some N).

N a

a

n a

Figure 2.11 An illustration of the formal definition of limits to show convergence of the sequenceantoa asn→ ∞: For alln>N,anlies in the strip of width 2 and centered at a.

EXAMPLE 9 Show that

n→∞lim 1 n =⁰

Solution

Before we show this for any arbitrary , let’s try to findN for a particular choice of , say, =⁰.03. We need to find an integerN such that

n¹ −⁰

<⁰.^{03 whenever}n>N Solving the inequality|_n¹−⁰|<⁰.^{03 for}npositive, we find that

¹n

<⁰.⁰³, ^or n> ¹

0.03 ≈³³.³³

The smallest value forN that we can choose isN =33, which is the largest integer less than or equal to 1/⁰.03. Successive values forn>33 give us confidence that we are on the right track but don't prove that our choice is correct:

a₃₄= ¹

34 ≈⁰.⁰²⁹⁴, a₃₅= ¹

35 ≈⁰.⁰²⁸⁶, ^{and so on}

To see that our choice forNworks, we need to show thatn>^{33 implies}|¹/n|<⁰.^03.

Now, sincen takes on only integer values,n > 33 is equivalent ton ≥ ^{34, which} implies that 1/n≤¹/³⁴≈⁰.0294. Sincen>^{0, we have}

n¹ −⁰

<⁰.^{03 whenever}n>³³

To show thatan = _n¹ converges to 0, we need to do the same calculation for any arbitrary . That is, we need to show that, for every >0, we can find anN such that

¹n −⁰

< ^whenevern> N

To find a candidate forN, we solve the inequality|¹_n|< ^{. Since 1}_n >0, we can drop the absolute-value signs and find

1

n < , ^or n> ¹

Let’s chooseN so that 1/N ≥ ^{and 1}/(N +¹) < , or, equivalently,N ≤¹/ ^and N+¹>¹/ . This means that we chooseNto be the largest integer less than or equal to 1/ ^{. If}n> N, thenn≥ N+1, which is equivalent to 1/n≤¹/(N +¹)^{. Since} N is the largest integer less than or equal to 1/ , it follows that 1/n≤¹/(N+¹) < ≤ 1/N forn > N. This condition, together withn > 0, shows that if N is the largest integer less than or equal to 1/ ^{, then}

¹n −⁰

< ^whenevern> N

Limit Laws The formal definition of limits is cumbersome when we want to compute limits in specific examples. Fortunately, there are mathematical laws that facilitate the computation of limits:

Limit Laws If lim_n→∞an and lim_n→∞bnexist andcis a constant, then

n→∞lim(an+bn)= ^lim

n→∞an + ^lim

n→∞bn

lim

n→∞(ca_n)=c lim

n→∞a_n

n→∞lim(a_nb_n)=(^lim

n→∞a_n)(^lim

n→∞b_n)

n→∞lim a_n

b_n = ^limn→∞a_n

lim_n→∞b_n, provided lim

n→∞b_n =⁰

Although we do not need to know the formal definition of limits in order to use the limit laws, in the next two examples we will need to know that

lim

n→∞

1

n =^{0 (2.6)}

which was proved (using the formal definition of limits) in Example 9.

EXAMPLE 10 Find

lim

n→∞

n+¹ n

Solution

We break ⁿ⁺¹_n into a sum of two terms, namely, 1+_n¹. Since lim_n→∞1 and lim_n→∞n¹ exist [the former is equal to 1 and the latter to 0, according to (2.6)], it follows that

n→∞lim n+¹

n = ^lim

n→∞

1+ ¹

n

= ^lim

n→∞(¹)+ ^lim

n→∞

1

n =¹+⁰=¹ as claimed in Example 8.

EXAMPLE 11 Find

n→∞lim

4n²−¹ n²

Solution

We rewritean:

a_n = ⁴n²−¹

n² =⁴− ¹

n² =⁴− ¹ n· ¹

n Since lim_n→∞4 and lim_n→∞n¹ exist, we have

lim

n→∞

4n²−1 n² = lim

n→∞

4− ¹

n· ¹ n

= lim

n→∞4−

lim

n→∞

1

n n^lim→∞

1 n

=⁴−⁰·⁰=⁴

EXAMPLE 12 Without proof, we will state the long-term behavior of exponential growth. ForR>

0, exponential growth is given by

a_n =a₀Rⁿ,n=⁰,¹,², . . . Figure 2.12 indicates that

n→∞lim an =

⎧⎪

⎨

⎪⎩

0 if 0<R<¹ a₀ ifR=¹

∞ ^ifR>¹

an

a0

0

0 2 4 6

n

8 10 12

0 R 1 R 1 R 1

Figure 2.12 Exponential growth in Example 12 for three different values ofR.

This conclusion can also be shown rigorously by using the formal definition of limits.