SOLUTIONS TO LIMITS OF FUNCTIONS USING THE PRECISE DEFINITION OF LIMIT



SOLUTION 1 :

Prove that tex2html_wrap_inline546 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 so that if tex2html_wrap_inline552 , then tex2html_wrap_inline554, i.e., tex2html_wrap_inline556 , i.e., tex2html_wrap_inline558 . But this trivial inequality is always true, no matter what value is chosen for tex2html_wrap_inline560. For example, tex2html_wrap_inline562 will work. Thus, if tex2html_wrap_inline552 , then it follows that tex2html_wrap_inline554 . This completes the proof.

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SOLUTION 2 : Prove that tex2html_wrap_inline568 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline576 , then tex2html_wrap_inline578. Begin with tex2html_wrap_inline578 and ``solve for" |x-10| . Then,

tex2html_wrap_inline578 iff tex2html_wrap_inline586

iff tex2html_wrap_inline588

iff tex2html_wrap_inline590

iff tex2html_wrap_inline592

iff tex2html_wrap_inline594.

Now choose tex2html_wrap_inline596 . Thus, if tex2html_wrap_inline598 , it follows that tex2html_wrap_inline578. This completes the proof.

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SOLUTION 3 : Prove that tex2html_wrap_inline602 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline610 , then tex2html_wrap_inline554. Begin with tex2html_wrap_inline554 and ``solve for" tex2html_wrap_inline616 . Then,

tex2html_wrap_inline554 iff tex2html_wrap_inline620

iff tex2html_wrap_inline622

iff tex2html_wrap_inline624

iff tex2html_wrap_inline626

iff tex2html_wrap_inline628

iff tex2html_wrap_inline630 .

Now choose tex2html_wrap_inline632 . Thus, if tex2html_wrap_inline634 , it follows that tex2html_wrap_inline554. This completes the proof.

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SOLUTION 4 : Prove that tex2html_wrap_inline638 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline646 , then tex2html_wrap_inline648. Begin with tex2html_wrap_inline648 and ``solve for" |x-1| . Then,

tex2html_wrap_inline648 iff tex2html_wrap_inline656

iff tex2html_wrap_inline658

iff tex2html_wrap_inline660

iff tex2html_wrap_inline662 .

We will now ``replace" the term |x+1| with an appropriate constant and keep the term |x-1| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that tex2html_wrap_inline668 (This is a valid assumption to make since, in general, once we find a tex2html_wrap_inline560 that works, all smaller values of tex2html_wrap_inline560 also work.) . Then tex2html_wrap_inline674 implies that -1 < x-1 < 1 and 0 < x < 2 so that 1 < |x+1| < 3 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline684

iff tex2html_wrap_inline686

iff tex2html_wrap_inline688 .

Now choose tex2html_wrap_inline690 (This guarantees that both assumptions made about tex2html_wrap_inline560 in the course of this proof are taken into account simultaneously.). Thus, if tex2html_wrap_inline646 , it follows that tex2html_wrap_inline648. This completes the proof.

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SOLUTION 5 : Prove that tex2html_wrap_inline698 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline706 , then tex2html_wrap_inline648. Begin with tex2html_wrap_inline648 and ``solve for" | x - (-1) | = | x + 1 | . Then,

tex2html_wrap_inline648 iff tex2html_wrap_inline656

iff tex2html_wrap_inline658

iff tex2html_wrap_inline660

iff tex2html_wrap_inline662 .

We will now ``replace" the term |x-1| with an appropriate constant and keep the term |x+1| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that tex2html_wrap_inline668 (This is a valid assumption to make since, in general, once we find a tex2html_wrap_inline560 that works, all smaller values of tex2html_wrap_inline560 also work.). Then tex2html_wrap_inline734 implies that -1 < x+1 < 1 and -2 < x < 0 so that 1 < |x-1| < 3 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline744

iff tex2html_wrap_inline746

iff tex2html_wrap_inline748 .

Now choose tex2html_wrap_inline690 (This guarantees that both assumptions made about tex2html_wrap_inline560 in the course of this proof are taken into account simultaneously.). Thus, if tex2html_wrap_inline754 , it follows that tex2html_wrap_inline648. This completes the proof.

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SOLUTION 6 : Prove that tex2html_wrap_inline758 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline766 , then tex2html_wrap_inline768. Begin with tex2html_wrap_inline768 and ``solve for" | x - 2 | . Then,

tex2html_wrap_inline768 iff tex2html_wrap_inline776

iff tex2html_wrap_inline778

iff tex2html_wrap_inline780 .

We will now ``replace" the term |3x+5| with an appropriate constant and keep the term |x-2| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that tex2html_wrap_inline668 (This is a valid assumption to make since, in general, once we find a tex2html_wrap_inline560 that works, all smaller values of tex2html_wrap_inline560 also work.) . Then tex2html_wrap_inline792 implies that -1 < x-2 < 1 and 1 < x < 3 so that 8 < |3x+5| < 14 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline802

iff tex2html_wrap_inline804

iff tex2html_wrap_inline806 .

Now choose tex2html_wrap_inline808 (This guarantees that both assumptions made about tex2html_wrap_inline560 in the course of this proof are taken into account simultaneously.). Thus, if tex2html_wrap_inline766 , it follows that tex2html_wrap_inline768. This completes the proof.

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SOLUTION 7 : Prove that tex2html_wrap_inline816. Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline824 , then tex2html_wrap_inline826. Begin with tex2html_wrap_inline826 and ``solve for" | x - 3 | . Then,

tex2html_wrap_inline826 iff tex2html_wrap_inline834

iff tex2html_wrap_inline836

iff tex2html_wrap_inline838

iff tex2html_wrap_inline840

iff tex2html_wrap_inline842

iff tex2html_wrap_inline844

iff tex2html_wrap_inline846

iff tex2html_wrap_inline848 .

We will now ``replace" the term |x+3| with an appropriate constant and keep the term |x-3| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that tex2html_wrap_inline668 (This is a valid assumption to make since, in general, once we find a tex2html_wrap_inline560 that works, all smaller values of tex2html_wrap_inline560 also work.) . Then tex2html_wrap_inline860 implies that -1 < x-3 < 1 and 2 < x < 4 so that 5 < |x+3| < 7 and tex2html_wrap_inline868 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline872

iff tex2html_wrap_inline874

iff tex2html_wrap_inline876

iff tex2html_wrap_inline878 .

Now choose tex2html_wrap_inline880 (This guarantees that both assumptions made about tex2html_wrap_inline560 in the course of this proof are taken into account simultaneously.). Thus, if tex2html_wrap_inline824 , it follows that tex2html_wrap_inline826. This completes the proof.

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SOLUTION 8 : Prove that tex2html_wrap_inline888 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline896 , then tex2html_wrap_inline898. Begin with tex2html_wrap_inline900 and ``solve for" | x - (-6) | = | x + 6 | . Then,

tex2html_wrap_inline904 iff tex2html_wrap_inline906

iff tex2html_wrap_inline908

iff tex2html_wrap_inline910

iff tex2html_wrap_inline912

iff tex2html_wrap_inline914

iff tex2html_wrap_inline916

iff tex2html_wrap_inline918

iff tex2html_wrap_inline920 .

We will now ``replace" the term |2-x| with an appropriate constant and keep the term |x+6| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that tex2html_wrap_inline668 (This is a valid assumption to make since, in general, once we find a tex2html_wrap_inline560 that works, all smaller values of tex2html_wrap_inline560 also work.) . Then tex2html_wrap_inline932 implies that -1 < x+6 < 1 and -7 < x < -5 so that 7 < |2-x| < 9 and tex2html_wrap_inline940 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline944 .

iff tex2html_wrap_inline946 .

iff tex2html_wrap_inline948

iff tex2html_wrap_inline950 .

Now choose tex2html_wrap_inline952 (This guarantees that both assumptions made about tex2html_wrap_inline560 in the course of this proof are taken into account simultaneously.). Thus, if tex2html_wrap_inline956 , it follows that tex2html_wrap_inline900. This completes the proof.

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SOLUTION 9 : Prove that tex2html_wrap_inline960 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline824 , then tex2html_wrap_inline970. Begin with tex2html_wrap_inline970 and ``solve for" | x - 3 | . Then,

tex2html_wrap_inline970 iff tex2html_wrap_inline978

iff tex2html_wrap_inline980

iff tex2html_wrap_inline982

iff tex2html_wrap_inline984

iff tex2html_wrap_inline986

iff tex2html_wrap_inline988

iff tex2html_wrap_inline990 .

We will now ``replace" the term | 4x-9 | with an appropriate constant and keep the term |x-3| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that tex2html_wrap_inline668 (This is a valid assumption to make since, in general, once we find a tex2html_wrap_inline560 that works, all smaller values of tex2html_wrap_inline560 also work.) . Then tex2html_wrap_inline860 implies that -1 < x-3 < 1 and 2 < x < 4 . HOWEVER, THIS RANGE OF X-VALUES IS NOT APPROPRIATE SINCE THE FUNCTION tex2html_wrap_inline1008 IS NOT DEFINED AT tex2html_wrap_inline1010 ! Fortunately, this problem can be easily resolved. We simply pick tex2html_wrap_inline560 small enough to avoid tex2html_wrap_inline1010 . For example, assume that tex2html_wrap_inline1016. Then tex2html_wrap_inline1018 implies that tex2html_wrap_inline1020 and tex2html_wrap_inline1022 so that 2 < | 4x-9 | < 4 and tex2html_wrap_inline1026 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline1030 .

iff tex2html_wrap_inline1032

iff tex2html_wrap_inline1034

iff tex2html_wrap_inline1036 .

Now choose tex2html_wrap_inline1038 (This guarantees that both assumptions made about tex2html_wrap_inline560 in the course of this proof are taken into account simultaneously.). Thus, if tex2html_wrap_inline824 , it follows that tex2html_wrap_inline970. This completes the proof.

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SOLUTION 10 : Prove that tex2html_wrap_inline1046 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline1054 , then tex2html_wrap_inline1056. Begin with tex2html_wrap_inline1056 and ``solve for" | x - 9 | . Then,

tex2html_wrap_inline1056 iff tex2html_wrap_inline1064

iff tex2html_wrap_inline1066

(At this point, we need to figure out a way to make | x-9 | ``appear'' in our computations. Appropriate use of the conjugate will suffice.)

iff tex2html_wrap_inline1070

(Recall that tex2html_wrap_inline1072 .)

iff tex2html_wrap_inline1074

iff tex2html_wrap_inline1076 .

iff tex2html_wrap_inline1078 .

We will now ``replace" the term tex2html_wrap_inline1080 with an appropriate constant and keep the term |x-9| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that tex2html_wrap_inline668 (This is a valid assumption to make since, in general, once we find a tex2html_wrap_inline560 that works, all smaller values of tex2html_wrap_inline560 also work.) . Then tex2html_wrap_inline1090 implies that -1 < x-9 < 1 and 8 < x < 10 so that tex2html_wrap_inline1096 and tex2html_wrap_inline1098 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline1102

iff tex2html_wrap_inline1104

iff tex2html_wrap_inline1106 .

Now choose tex2html_wrap_inline1108 (This guarantees that both assumptions made about tex2html_wrap_inline560 in the course of this proof are taken into account simultaneously.). Thus, if tex2html_wrap_inline1054 , it follows that tex2html_wrap_inline1056. This completes the proof.

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SOLUTION 11 : Prove that tex2html_wrap_inline1116 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline1124 , then tex2html_wrap_inline1126. Begin with tex2html_wrap_inline1126 and ``solve for" | x - 4 | . Then,

tex2html_wrap_inline1126 iff tex2html_wrap_inline1134

(At this point, we need to figure out a way to make | x-4 | ``appear'' in our computations. Appropriate use of the conjugate will suffice.)

iff tex2html_wrap_inline1138

(Recall that tex2html_wrap_inline1072 .)

iff tex2html_wrap_inline1142

iff tex2html_wrap_inline1144

iff tex2html_wrap_inline1146

iff tex2html_wrap_inline1148 .

We will now ``replace" the term tex2html_wrap_inline1150 with an appropriate constant and keep the term |x-4| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that tex2html_wrap_inline668 (This is a valid assumption to make since, in general, once we find a tex2html_wrap_inline560 that works, all smaller values of tex2html_wrap_inline560 also work.) . Then tex2html_wrap_inline1160 implies that -1 < x-4 < 1 and 3 < x < 5 so that tex2html_wrap_inline1166 and tex2html_wrap_inline1168 (Make sure that you understand this step before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline1172

iff tex2html_wrap_inline1174

iff tex2html_wrap_inline1176 .

Now choose tex2html_wrap_inline1108 (This guarantees that both assumptions made about tex2html_wrap_inline560 in the course of this proof are taken into account simultaneously.). Thus, if tex2html_wrap_inline1124 , it follows that tex2html_wrap_inline1126. This completes the proof.

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SOLUTION 12 : Prove that tex2html_wrap_inline1186 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline646 , then tex2html_wrap_inline1196. Begin with tex2html_wrap_inline1196 and ``solve for" | x - 1 | . Then,

tex2html_wrap_inline1196 iff tex2html_wrap_inline1204

iff tex2html_wrap_inline1206

iff tex2html_wrap_inline1208

iff tex2html_wrap_inline1210

iff tex2html_wrap_inline1212

(At this point, we need to figure out a way to make | x-1 | ``appear'' in our computations. A simple use of constants will get us started.)

iff tex2html_wrap_inline1216

iff tex2html_wrap_inline1218

iff tex2html_wrap_inline1220

(We need to be able to factor (x-1) from the numerator. Apply the conjugate to the term tex2html_wrap_inline1224.)

iff tex2html_wrap_inline1226

iff tex2html_wrap_inline1228

iff tex2html_wrap_inline1230

iff tex2html_wrap_inline1232

(Now get a common denominator.)

iff tex2html_wrap_inline1234

iff tex2html_wrap_inline1236

iff tex2html_wrap_inline1238

iff tex2html_wrap_inline1240

iff tex2html_wrap_inline1242 .

We will now ``replace" the terms tex2html_wrap_inline1244 and tex2html_wrap_inline1246 with appropriate constants and keep the term |x-1| , since this is the term we wish to ``solve for". To do this, we will arbitrarily assume that tex2html_wrap_inline668 (This is a valid assumption to make since, in general, once we find a tex2html_wrap_inline560 that works, all smaller values of tex2html_wrap_inline560 also work.). Then tex2html_wrap_inline674 implies that -1 < x-1 < 1 and 0 < x < 2 so that tex2html_wrap_inline1262 . In addition, tex2html_wrap_inline1264 so that tex2html_wrap_inline1266 and tex2html_wrap_inline1268 (Make sure that you understand these steps before proceeding.). It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline1272

iff tex2html_wrap_inline1274

iff tex2html_wrap_inline1276 .

Now choose tex2html_wrap_inline1278 (This guarantees that both assumptions made about tex2html_wrap_inline560 in the course of this proof are taken into account simultaneously.). Thus, if tex2html_wrap_inline646 , it follows that tex2html_wrap_inline1196. This completes the proof.

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SOLUTION 13 : Prove that tex2html_wrap_inline1286 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline1294 , then tex2html_wrap_inline1296. Begin with tex2html_wrap_inline1296 and ``solve for" | x - a | . Then,

tex2html_wrap_inline1296 iff tex2html_wrap_inline1304 .

At this point, we need to figure out a way to introduce the term | x-a | into our computations. The answer lies with the Mean Value Theorem. Consider the function tex2html_wrap_inline1308 on the interval [A, B] . Since f is continuous on the closed interval [A, B] and differentiable ( tex2html_wrap_inline1316 ) on the open interval (A, B) , according to the Mean Value Theorem there is at least one number C , A < C < B , satisfying

tex2html_wrap_inline1324 ,

i.e.,

tex2html_wrap_inline1326.

Then

tex2html_wrap_inline1328

so that

tex2html_wrap_inline1330.

This is true for any two real numbers, A and B . It follows that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline1338 .

Now choose tex2html_wrap_inline1340 . Thus, if tex2html_wrap_inline1294 , it follows that tex2html_wrap_inline1296. This completes the proof.

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SOLUTION 14 : Prove that tex2html_wrap_inline1346 . Begin by letting tex2html_wrap_inline548 be given. Find tex2html_wrap_inline550 (which depends on tex2html_wrap_inline574 ) so that if tex2html_wrap_inline1294 , then tex2html_wrap_inline1356. Begin with tex2html_wrap_inline1356 and ``solve for" | x - a | . Then,

tex2html_wrap_inline1356 iff tex2html_wrap_inline1364 .

At this point, we need to figure out a way to introduce the term | x-a | into our computations. The answer lies with the Mean Value Theorem. Consider the function tex2html_wrap_inline1368 on the interval [A, B] , where A and B are both positive. Since f is continuous on the closed interval [A, B] and differentiable ( tex2html_wrap_inline1380 ) on the open interval (A, B) , according to the Mean Value Theorem there is at least one number C , A < C < B , satisfying

tex2html_wrap_inline1324 ,

i.e.,

tex2html_wrap_inline1390.

Then, since 0 < A < C < B , it follows that tex2html_wrap_inline1394 , so that

tex2html_wrap_inline1396 .

Thus,

(*) tex2html_wrap_inline1398.

This is true for any two positive real numbers A and B , where B > A . At this point, we need to consider two cases.

If x > a , it follows from inequality (*) that (Always make this ``replacement" between your last expression on the left and tex2html_wrap_inline574. This guarantees the logic of the proof.)

tex2html_wrap_inline1410

iff tex2html_wrap_inline1412

iff tex2html_wrap_inline1414 .

Now choose tex2html_wrap_inline1416 . Thus, if tex2html_wrap_inline1294 , it follows that tex2html_wrap_inline1356.

If x < a , then it is reasonable to assume that tex2html_wrap_inline1424 since we are considering the limit as x approaches a . Thus, tex2html_wrap_inline1430 and it follows from inequality (*) that

tex2html_wrap_inline1432

so that

tex2html_wrap_inline1434

iff tex2html_wrap_inline1436

iff tex2html_wrap_inline1438 .

Now choose tex2html_wrap_inline1440 . Thus, if tex2html_wrap_inline1294 , it follows that tex2html_wrap_inline1356. This completes the proof.

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SOLUTION 15 : Let tex2html_wrap_inline1446 . Prove that tex2html_wrap_inline1448 does not exist .

ASSUME THAT THE LIMIT DOES EXIST. That is, assume that tex2html_wrap_inline1450 , where L is some real number. It follows that for EACH real number tex2html_wrap_inline548 , there exists another real number tex2html_wrap_inline550 so that

if tex2html_wrap_inline646 , then tex2html_wrap_inline1460 .

We will proceed to find ONE tex2html_wrap_inline548 for which NO tex2html_wrap_inline550 works. THIS WILL BE A CONTRADICTION OF OUR ASSUMPTION, making our assumption false, proving that the limit does not exist. (This method is called proof by contradiction.)





By looking at the graph of f , which is given above, we see that x-values chosen ``near'' to x=1 but on opposite sides of x=1 have corresponding y-values which are ``about'' one unit apart. Intuitively, this tells us that the limit does not exist and leads us to choose an appropriate tex2html_wrap_inline548 leading to the above contradiction.

Consider tex2html_wrap_inline1478. Under our assumption that the limit does exist, it follows that there is some number tex2html_wrap_inline550 so that if tex2html_wrap_inline646, then tex2html_wrap_inline1484 . But for ANY choice of tex2html_wrap_inline550

tex2html_wrap_inline646 iff tex2html_wrap_inline1490 (and x not equal to 1)

iff tex2html_wrap_inline1492 (and x not equal to 1) .

Thus, both tex2html_wrap_inline1494 and tex2html_wrap_inline1496 satisfy tex2html_wrap_inline646, which implies that

tex2html_wrap_inline1500 and tex2html_wrap_inline1502.

In addition,

tex2html_wrap_inline1504

tex2html_wrap_inline1506

tex2html_wrap_inline1508

tex2html_wrap_inline1510

= | -1 |

= 1 .

Now, by the triangle inequality

tex2html_wrap_inline1516

tex2html_wrap_inline1518

tex2html_wrap_inline1520

tex2html_wrap_inline1522

tex2html_wrap_inline1524

= 1 .

We have just concluded that

tex2html_wrap_inline1528 ,

an obvious contradiction. It must be that the original limit does not exist.

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Duane Kouba
Tue May 6 10:38:27 PDT 1997