The previous discussion should have conveyed some general ideas about the limit concept. Let us now give it a more precise definition. Since such a definition will make use of the concept of neighborhood of a point on a line (in particular, a specific number as a point on the line of real numbers), we shall first explain the latter term.

For a given number ¿, there can always be found a number (L — a\ ) < L and another number (L+ ai) > where u\ and a2 are some arbitrary positive numbers. The set of all numbers falling between {L -a|) and (L + a2) is called the interval between those two numbers. If the numbers (L — a j) and (L + a2) are included in the set, the set is a closed interval; if they are excluded, the set is an open interval. A closed interval between (L - a{) and (L + a2) is denoted by the bracketed expression

[L - ti{, L +a2] = {q \ L - ay < q < L + a2] and the corresponding open interval is denoted with parentheses:

(L - tfj, L +¿72) = [q I - <q < L + a2) (6,4)

Thus, [ ] relate to the weak inequality sign <, whereas ( ) relate to the strict inequality sign <> But in both types of intervals, the smaller number (L - a}) is always listed first. Later on, we shall also have occasion to refer to half-open and half-closed intervals such as (3, 5] and f6, oo), which have the following meanings:

(3, 5] = {x \ 3 <x <5} [6, oo) = {x | 6 < x < oo|

Now we may define a neighborhood of L to be an open interval as defined in (6.4), which is an interval "covering" the number L.+ Depending on the magnitudes of the arbitrary numbers a\ and ai, it is possible to construct various neighborhoods for the given number i,t Using the conccpt of neighborhood, the limit of a function may then be defined as follows:

As t> approaches a number Ny the limit of q = g{ v) is the number L, if, for every neighborhood of i that can be choscn, however small there can he found a corresponding neighborhood of N (excluding the point v = N) in the domain of the function such thai, for every value of v in that N-neighborhood, its image lies in the chosen ¿-neighborhood.

This statement can be clarified with the help of Fig. 6.3, which resembles Fig. 6.2a. From what was learned about Fig. 6.2th we know that lim q — L in Fig. 6.3. Let us show that L does indeed fulfill the new definition of a limit. As the first step, select an arbitrary small neighborhood of say, (L — a^, L + a2). (This should have bct?n made even smaller, but we are keeping it relatively large to facilitate exposition.) Now construct a neighborhood of N, say, (N - bu N + h2), such that the two neighborhoods (when extended into quadrant I) will together define a rectangle (shaded in diagram) with two of its corners lying on the given curve. It can then be verified that, for every value of v in this neighborhood of N (not counting v = N\ the corresponding value of q = g(v) lies in the

* The identification of an open interval as the neighborhood of a point is valid only when we are considering a point on a line (one-dimensional space). In the case of a point in a plane (twQ-dimensionaf space), its neighborhood must be thought of as an area, say, a circular area that includes the point.

FIGURE 63

FIGURE 63

chosen neighborhood of L. In fact, no matter how mail an ¿-neighborhood we choose, a (correspondingly small) /V-neighborhood can be found with the property jusl cited. Thus L fulfills the definition of a limit, as was to be demonstrated.

We can also apply the given definition to the step function of Fig. 6.2c in order to show that neither ¿i nor L2 qualifies as lim qt If wC choose a very small neighborhood of L\ —

say Just a hair's width on each side of L \—then, no matter what neighborhood we pick for N, the rectangle associated with the two neighborhoods cannot possibly enclose the lower step of the function. Consequently, for any value of v > N, the corresponding value of q (located on Ihe lower step) will not be in the neighborhood of L [ ? and thus i \ fails the test for a limit. By similar reasoning, L2 must also be dismissed as a candidate for lim q. In fact, in this case no limit exists for q as v N. ' ^

The fulfillment of the definition can also be chccked algebraically rather than by graph. For instance, consider again the function

It has been found in Example 2 that lim q = 2; thus, here we have N = I and 1=2. To l' - > 1

verify that L — 2 is indeed the limit of q, we must demonstrate that, for every chosen neighborhood of ¿, (2 - 2 + a2), there exists a neighborhood of /V, (1 - h\, 1 + h2), such that, whenever v is in this neighborhood of N, q must be in the chosen neighborhood of L. This means essentially that, for given values of at and a2, however small, two numbers b [ and b2 must be found such that, whenever the inequality

1 - hx < v < I -f- h2 (MO is satisfied, another inequality of the form

must also be satisfied. To find such a pair of numbers h\ and hi, let us first rewrite (6.7) by substituting (6.5):

2 -a\ < I + i' < 2 + a7 (6.7') This, in turn, can be transformed (by subtracting 1 from each side) into the inequality

A comparison of (6.7")—a variant of (6.7) with (6,6) suggests that if we choose the two numbers b\ and hi to be b\ = ct\ and b2 — a2> the two inequalities (6.6) and (6.7) will always be satisfied simultaneously. Thus the neighborhood of N\ (1 — b\, I -f-hi), as required in the definition of a limit, can indeed be found for the case of /. = 2. and ihis establishes L = 2 as the limit.

Let us now utilize the definition of a limit in the opposite way, to show that another value (say, 3) cannot qualify as lim q for the function in (6.5). If 3 were that limit, it would have

to be true that, for every chosen neighborhood of 3, (3 - uu 3 + az), there exists a neighborhood of 1,(1 - b\, 1 + ¿2), such that, whenever v is in the tatter neighborhood, q must be in the former neighborhood. That is, whenever the inequality

1 — b\ < v < 1 + h2 is satisfied, another inequality of the form

3 — u\ < 1 4- i1 < 3 -f a2 or 2 - <7j < v < 2 + «2

must also be satisfied. The only way to achieve this result is to choose h\ = a\ - I and hi — a2 + 1. This would imply that the neighborhood of 1 is to be the open interval (2 - r/i, 2 + a2). According to the definition of a limit, however, a \ and g-± can be made arbitrarily small, say, a\ — a2 = 0.1, In that case, the last-mentioned interval will turn out to be (1,9,2,1) which lies entirely to the right of the point v ~ 1 on the horizontal axis and, hcncc, does not even qualify as a neighborhood of 1. Thus the definition of a limit cannot be satisfied by the number 3. A similar procedure can be employed lo show that any number other than 2 will contradict the definition of a limit in the present case.

In general, if one number satisfies the definition of a limit of q as v —1► A\ then no other number can. If a limit exists, it is unique.

1. Given the function q = (v2 + v- 56)/(v- 7), (v^ 7), find the left-side limit and the right-side limit of qas vapproaches 7. Can we conclude from these answers that q has a limit as v approaches 11

4. Use Fig. 6,3 to show that we cannot consider the number (i + 02) as the limit of q as v tends to N.

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