report

ap calculus 2.1 homework

lim x โ†’ 1 1 x โˆ’ 1 x โˆ’ 1 = โˆ’1 lim x โ†’ 1 1 x โˆ’ 1 x โˆ’ 1 = โˆ’1

lim x โ†’ 2 h ( x ) = โˆ’1 . lim x โ†’ 2 h ( x ) = โˆ’1 .

lim x โ†’ 2 | x 2 โˆ’ 4 | x โˆ’ 2 lim x โ†’ 2 | x 2 โˆ’ 4 | x โˆ’ 2 does not exist.

a. lim x โ†’ 2 โˆ’ | x 2 โˆ’ 4 | x โˆ’ 2 = โˆ’4 ; lim x โ†’ 2 โˆ’ | x 2 โˆ’ 4 | x โˆ’ 2 = โˆ’4 ; b. lim x โ†’ 2 + | x 2 โˆ’ 4 | x โˆ’ 2 = 4 lim x โ†’ 2 + | x 2 โˆ’ 4 | x โˆ’ 2 = 4

a. lim x โ†’ 0 โˆ’ 1 x 2 = + โˆž ; lim x โ†’ 0 โˆ’ 1 x 2 = + โˆž ; b. lim x โ†’ 0 + 1 x 2 = + โˆž ; lim x โ†’ 0 + 1 x 2 = + โˆž ; c. lim x โ†’ 0 1 x 2 = + โˆž lim x โ†’ 0 1 x 2 = + โˆž

a. lim x โ†’ 2 โˆ’ 1 ( x โˆ’ 2 ) 3 = โˆ’ โˆž ; lim x โ†’ 2 โˆ’ 1 ( x โˆ’ 2 ) 3 = โˆ’ โˆž ; b. lim x โ†’ 2 + 1 ( x โˆ’ 2 ) 3 = + โˆž ; lim x โ†’ 2 + 1 ( x โˆ’ 2 ) 3 = + โˆž ; c. lim x โ†’ 2 1 ( x โˆ’ 2 ) 3 lim x โ†’ 2 1 ( x โˆ’ 2 ) 3 DNE. The line x = 2 x = 2 is the vertical asymptote of f ( x ) = 1 / ( x โˆ’ 2 ) 3 . f ( x ) = 1 / ( x โˆ’ 2 ) 3 .

Does not exist.

11 10 11 10

lim x โ†’ โˆ’1 โˆ’ f ( x ) = โˆ’1 lim x โ†’ โˆ’1 โˆ’ f ( x ) = โˆ’1

f is not continuous at 1 because f ( 1 ) = 2 โ‰  3 = lim x โ†’ 1 f ( x ) . f ( 1 ) = 2 โ‰  3 = lim x โ†’ 1 f ( x ) .

f ( x ) f ( x ) is continuous at every real number.

Discontinuous at 1; removable

[ โˆ’3 , + โˆž ) [ โˆ’3 , + โˆž )

f ( 0 ) = 1 > 0 , f ( 1 ) = โˆ’2 < 0 ; f ( x ) f ( 0 ) = 1 > 0 , f ( 1 ) = โˆ’2 < 0 ; f ( x ) is continuous over [ 0 , 1 ] . [ 0 , 1 ] . It must have a zero on this interval.

Let ฮต > 0 ; ฮต > 0 ; choose ฮด = ฮต 3 ; ฮด = ฮต 3 ; assume 0 < | x โˆ’ 2 | < ฮด . 0 < | x โˆ’ 2 | < ฮด .

Thus, | ( 3 x โˆ’ 2 ) โˆ’ 4 | = | 3 x โˆ’ 6 | = | 3 | ยท | x โˆ’ 2 | < 3 ยท ฮด = 3 ยท ( ฮต / 3 ) = ฮต . | ( 3 x โˆ’ 2 ) โˆ’ 4 | = | 3 x โˆ’ 6 | = | 3 | ยท | x โˆ’ 2 | < 3 ยท ฮด = 3 ยท ( ฮต / 3 ) = ฮต .

Therefore, lim x โ†’ 2 3 x โˆ’ 2 = 4 . lim x โ†’ 2 3 x โˆ’ 2 = 4 .

Choose ฮด = min { 9 โˆ’ ( 3 โˆ’ ฮต ) 2 , ( 3 + ฮต ) 2 โˆ’ 9 } . ฮด = min { 9 โˆ’ ( 3 โˆ’ ฮต ) 2 , ( 3 + ฮต ) 2 โˆ’ 9 } .

| x 2 โˆ’ 1 | = | x โˆ’ 1 | ยท | x + 1 | < ฮต / 3 ยท 3 = ฮต | x 2 โˆ’ 1 | = | x โˆ’ 1 | ยท | x + 1 | < ฮต / 3 ยท 3 = ฮต

ฮด = ฮต 2 ฮด = ฮต 2

Section 2.1 Exercises

a. 2.2100000; b. 2.0201000; c. 2.0020010; d. 2.0002000; e. (1.1000000, 2.2100000); f. (1.0100000, 2.0201000); g. (1.0010000, 2.0020010); h. (1.0001000, 2.0002000); i. 2.1000000; j. 2.0100000; k. 2.0010000; l. 2.0001000

y = 2 x y = 2 x

a. 2.0248457; b. 2.0024984; c. 2.0002500; d. 2.0000250; e. (4.1000000,2.0248457); f. (4.0100000,2.0024984); g. (4.0010000,2.0002500); h. (4.00010000,2.0000250); i. 0.24845673; j. 0.24984395; k. 0.24998438; l. 0.24999844

y = x 4 + 1 y = x 4 + 1

a. โˆ’0.95238095; b. โˆ’0.99009901; c. โˆ’0.99502488; d. โˆ’0.99900100; e. (โˆ’1;.0500000,โˆ’0;.95238095); f. (โˆ’1;.0100000,โˆ’0;.9909901); g. (โˆ’1;.0050000,โˆ’0;.99502488); h. (1.0010000,โˆ’0;.99900100); i. โˆ’0.95238095; j. โˆ’0.99009901; k. โˆ’0.99502488; l. โˆ’0.99900100

y = โˆ’ x โˆ’ 2 y = โˆ’ x โˆ’ 2

โˆ’49 m/sec (velocity of the ball is 49 m/sec downward)

Under, 1 unit 2 ; over: 4 unit 2 . The exact area of the two triangles is 1 2 ( 1 ) ( 1 ) + 1 2 ( 2 ) ( 2 ) = 2.5 units 2 . 1 2 ( 1 ) ( 1 ) + 1 2 ( 2 ) ( 2 ) = 2.5 units 2 .

Under, 0.96 unit 2 ; over, 1.92 unit 2 . The exact area of the semicircle with radius 1 is ฯ€ ( 1 ) 2 2 = ฯ€ 2 ฯ€ ( 1 ) 2 2 = ฯ€ 2 unit 2 .

Approximately 1.3333333 unit 2

Section 2.2 Exercises

lim x โ†’ 1 f ( x ) lim x โ†’ 1 f ( x ) does not exist because lim x โ†’ 1 โˆ’ f ( x ) = โˆ’2 โ‰  lim x โ†’ 1 + f ( x ) = 2 . lim x โ†’ 1 โˆ’ f ( x ) = โˆ’2 โ‰  lim x โ†’ 1 + f ( x ) = 2 .

lim x โ†’ 0 ( 1 + x ) 1 / x = 2.7183 lim x โ†’ 0 ( 1 + x ) 1 / x = 2.7183

a. 1.98669331; b. 1.99986667; c. 1.99999867; d. 1.99999999; e. 1.98669331; f. 1.99986667; g. 1.99999867; h. 1.99999999; lim x โ†’ 0 sin 2 x x = 2 lim x โ†’ 0 sin 2 x x = 2

lim x โ†’ 0 sin a x x = a lim x โ†’ 0 sin a x x = a

a. โˆ’0.80000000; b. โˆ’0.98000000; c. โˆ’0.99800000; d. โˆ’0.99980000; e. โˆ’1.2000000; f. โˆ’1.0200000; g. โˆ’1.0020000; h. โˆ’1.0002000; lim x โ†’ 1 ( 1 โˆ’ 2 x ) = โˆ’1 lim x โ†’ 1 ( 1 โˆ’ 2 x ) = โˆ’1

a. โˆ’37.931934; b. โˆ’3377.9264; c. โˆ’333,777.93; d. โˆ’33,337,778; e. โˆ’29.032258; f. โˆ’3289.0365; g. โˆ’332,889.04; h. โˆ’33,328,889 lim x โ†’ 0 z โˆ’ 1 z 2 ( z + 3 ) = โˆ’ โˆž lim x โ†’ 0 z โˆ’ 1 z 2 ( z + 3 ) = โˆ’ โˆž

a. 0.13495277; b. 0.12594300; c. 0.12509381; d. 0.12500938; e. 0.11614402; f. 0.12406794; g. 0.12490631; h. 0.12499063; โˆด lim x โ†’ 2 1 โˆ’ 2 x x 2 โˆ’ 4 = 0.1250 = 1 8 โˆด lim x โ†’ 2 1 โˆ’ 2 x x 2 โˆ’ 4 = 0.1250 = 1 8

a. 10.00000; b. 100.00000; c. 1000.0000; d. 10,000.000; Guess: lim ฮฑ โ†’ 0 + 1 ฮฑ cos ( ฯ€ ฮฑ ) = โˆž , lim ฮฑ โ†’ 0 + 1 ฮฑ cos ( ฯ€ ฮฑ ) = โˆž , actual: DNE

False; lim x โ†’ โˆ’2 + f ( x ) = + โˆž lim x โ†’ โˆ’2 + f ( x ) = + โˆž

False; lim x โ†’ 6 f ( x ) lim x โ†’ 6 f ( x ) DNE since lim x โ†’ 6 โˆ’ f ( x ) = 2 lim x โ†’ 6 โˆ’ f ( x ) = 2 and lim x โ†’ 6 + f ( x ) = 5 . lim x โ†’ 6 + f ( x ) = 5 .

Answers may vary.

a. ฯ 2 ฯ 2 b. ฯ 1 ฯ 1 c. DNE unless ฯ 1 = ฯ 2 . ฯ 1 = ฯ 2 . As you approach x SF x SF from the left, you are in the high-density area of the shock. When you approach from the right, you have not experienced the โ€œshockโ€ yet and are at a lower density.

Section 2.3 Exercises

Use constant multiple law and difference law: lim x โ†’ 0 ( 4 x 2 โˆ’ 2 x + 3 ) = 4 lim x โ†’ 0 x 2 โˆ’ 2 lim x โ†’ 0 x + lim x โ†’ 0 3 = 3 lim x โ†’ 0 ( 4 x 2 โˆ’ 2 x + 3 ) = 4 lim x โ†’ 0 x 2 โˆ’ 2 lim x โ†’ 0 x + lim x โ†’ 0 3 = 3

Use root law: lim x โ†’ โˆ’2 x 2 โˆ’ 6 x + 3 = lim x โ†’ โˆ’2 ( x 2 โˆ’ 6 x + 3 ) = 19 lim x โ†’ โˆ’2 x 2 โˆ’ 6 x + 3 = lim x โ†’ โˆ’2 ( x 2 โˆ’ 6 x + 3 ) = 19

โˆ’ 5 7 โˆ’ 5 7

lim x โ†’ 4 x 2 โˆ’ 16 x โˆ’ 4 = 16 โˆ’ 16 4 โˆ’ 4 = 0 0 ; lim x โ†’ 4 x 2 โˆ’ 16 x โˆ’ 4 = 16 โˆ’ 16 4 โˆ’ 4 = 0 0 ; then, lim x โ†’ 4 x 2 โˆ’ 16 x โˆ’ 4 = lim x โ†’ 4 ( x + 4 ) ( x โˆ’ 4 ) x โˆ’ 4 = 8 lim x โ†’ 4 x 2 โˆ’ 16 x โˆ’ 4 = lim x โ†’ 4 ( x + 4 ) ( x โˆ’ 4 ) x โˆ’ 4 = 8

lim x โ†’ 6 3 x โˆ’ 18 2 x โˆ’ 12 = 18 โˆ’ 18 12 โˆ’ 12 = 0 0 ; lim x โ†’ 6 3 x โˆ’ 18 2 x โˆ’ 12 = 18 โˆ’ 18 12 โˆ’ 12 = 0 0 ; then, lim x โ†’ 6 3 x โˆ’ 18 2 x โˆ’ 12 = lim x โ†’ 6 3 ( x โˆ’ 6 ) 2 ( x โˆ’ 6 ) = 3 2 lim x โ†’ 6 3 x โˆ’ 18 2 x โˆ’ 12 = lim x โ†’ 6 3 ( x โˆ’ 6 ) 2 ( x โˆ’ 6 ) = 3 2

lim x โ†’ 9 t โˆ’ 9 t โˆ’ 3 = 9 โˆ’ 9 3 โˆ’ 3 = 0 0 ; lim x โ†’ 9 t โˆ’ 9 t โˆ’ 3 = 9 โˆ’ 9 3 โˆ’ 3 = 0 0 ; then, lim t โ†’ 9 t โˆ’ 9 t โˆ’ 3 = lim t โ†’ 9 t โˆ’ 9 t โˆ’ 3 t + 3 t + 3 = lim t โ†’ 9 ( t + 3 ) = 6 lim t โ†’ 9 t โˆ’ 9 t โˆ’ 3 = lim t โ†’ 9 t โˆ’ 9 t โˆ’ 3 t + 3 t + 3 = lim t โ†’ 9 ( t + 3 ) = 6

lim ฮธ โ†’ ฯ€ sin ฮธ tan ฮธ = sin ฯ€ tan ฯ€ = 0 0 ; lim ฮธ โ†’ ฯ€ sin ฮธ tan ฮธ = sin ฯ€ tan ฯ€ = 0 0 ; then, lim ฮธ โ†’ ฯ€ sin ฮธ tan ฮธ = lim ฮธ โ†’ ฯ€ sin ฮธ sin ฮธ cos ฮธ = lim ฮธ โ†’ ฯ€ cos ฮธ = โˆ’1 lim ฮธ โ†’ ฯ€ sin ฮธ tan ฮธ = lim ฮธ โ†’ ฯ€ sin ฮธ sin ฮธ cos ฮธ = lim ฮธ โ†’ ฯ€ cos ฮธ = โˆ’1

lim x โ†’ 1 / 2 2 x 2 + 3 x โˆ’ 2 2 x โˆ’ 1 = 1 2 + 3 2 โˆ’ 2 1 โˆ’ 1 = 0 0 ; lim x โ†’ 1 / 2 2 x 2 + 3 x โˆ’ 2 2 x โˆ’ 1 = 1 2 + 3 2 โˆ’ 2 1 โˆ’ 1 = 0 0 ; then, lim x โ†’ 1 / 2 2 x 2 + 3 x โˆ’ 2 2 x โˆ’ 1 = lim x โ†’ 1 / 2 ( 2 x โˆ’ 1 ) ( x + 2 ) 2 x โˆ’ 1 = 5 2 lim x โ†’ 1 / 2 2 x 2 + 3 x โˆ’ 2 2 x โˆ’ 1 = lim x โ†’ 1 / 2 ( 2 x โˆ’ 1 ) ( x + 2 ) 2 x โˆ’ 1 = 5 2

lim x โ†’ 6 2 f ( x ) g ( x ) = 2 lim x โ†’ 6 f ( x ) lim x โ†’ 6 g ( x ) = 72 lim x โ†’ 6 2 f ( x ) g ( x ) = 2 lim x โ†’ 6 f ( x ) lim x โ†’ 6 g ( x ) = 72

lim x โ†’ 6 ( f ( x ) + 1 3 g ( x ) ) = lim x โ†’ 6 f ( x ) + 1 3 lim x โ†’ 6 g ( x ) = 7 lim x โ†’ 6 ( f ( x ) + 1 3 g ( x ) ) = lim x โ†’ 6 f ( x ) + 1 3 lim x โ†’ 6 g ( x ) = 7

lim x โ†’ 6 g ( x ) โˆ’ f ( x ) = lim x โ†’ 6 g ( x ) โˆ’ lim x โ†’ 6 f ( x ) = 5 lim x โ†’ 6 g ( x ) โˆ’ f ( x ) = lim x โ†’ 6 g ( x ) โˆ’ lim x โ†’ 6 f ( x ) = 5

lim x โ†’ 6 [ ( x + 1 ) f ( x ) ] = ( lim x โ†’ 6 ( x + 1 ) ) ( lim x โ†’ 6 f ( x ) ) = 28 lim x โ†’ 6 [ ( x + 1 ) f ( x ) ] = ( lim x โ†’ 6 ( x + 1 ) ) ( lim x โ†’ 6 f ( x ) ) = 28

lim x โ†’ โˆ’3 โˆ’ ( f ( x ) โˆ’ 3 g ( x ) ) = lim x โ†’ โˆ’3 โˆ’ f ( x ) โˆ’ 3 lim x โ†’ โˆ’3 โˆ’ g ( x ) = 0 + 6 = 6 lim x โ†’ โˆ’3 โˆ’ ( f ( x ) โˆ’ 3 g ( x ) ) = lim x โ†’ โˆ’3 โˆ’ f ( x ) โˆ’ 3 lim x โ†’ โˆ’3 โˆ’ g ( x ) = 0 + 6 = 6

lim x โ†’ โˆ’5 2 + g ( x ) f ( x ) = 2 + ( lim x โ†’ โˆ’5 g ( x ) ) lim x โ†’ โˆ’5 f ( x ) = 2 + 0 2 = 1 lim x โ†’ โˆ’5 2 + g ( x ) f ( x ) = 2 + ( lim x โ†’ โˆ’5 g ( x ) ) lim x โ†’ โˆ’5 f ( x ) = 2 + 0 2 = 1

lim x โ†’ 1 f ( x ) โˆ’ g ( x ) 3 = lim x โ†’ 1 f ( x ) โˆ’ lim x โ†’ 1 g ( x ) 3 = 2 + 5 3 = 7 3 lim x โ†’ 1 f ( x ) โˆ’ g ( x ) 3 = lim x โ†’ 1 f ( x ) โˆ’ lim x โ†’ 1 g ( x ) 3 = 2 + 5 3 = 7 3

lim x โ†’ โˆ’9 ( x f ( x ) + 2 g ( x ) ) = ( lim x โ†’ โˆ’9 x ) ( lim x โ†’ โˆ’9 f ( x ) ) + 2 lim x โ†’ โˆ’9 ( g ( x ) ) = ( โˆ’9 ) ( 6 ) + 2 ( 4 ) = โˆ’46 lim x โ†’ โˆ’9 ( x f ( x ) + 2 g ( x ) ) = ( lim x โ†’ โˆ’9 x ) ( lim x โ†’ โˆ’9 f ( x ) ) + 2 lim x โ†’ โˆ’9 ( g ( x ) ) = ( โˆ’9 ) ( 6 ) + 2 ( 4 ) = โˆ’46

The limit is zero.

b. โˆž. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate negative distance.

Section 2.4 Exercises

The function is defined for all x in the interval ( 0 , โˆž ) . ( 0 , โˆž ) .

Removable discontinuity at x = 0 ; x = 0 ; infinite discontinuity at x = 1 x = 1

Infinite discontinuity at x = ln 2 x = ln 2

Infinite discontinuities at x = ( 2 k + 1 ) ฯ€ 4 , x = ( 2 k + 1 ) ฯ€ 4 , for k = 0 , ยฑ 1 , ยฑ 2 , ยฑ 3 ,โ€ฆ k = 0 , ยฑ 1 , ยฑ 2 , ยฑ 3 ,โ€ฆ

No. It is a removable discontinuity.

Yes. It is continuous.

k = โˆ’5 k = โˆ’5

k = โˆ’1 k = โˆ’1

k = 16 3 k = 16 3

Since both s and y = t y = t are continuous everywhere, then h ( t ) = s ( t ) โˆ’ t h ( t ) = s ( t ) โˆ’ t is continuous everywhere and, in particular, it is continuous over the closed interval [ 2 , 5 ] . [ 2 , 5 ] . Also, h ( 2 ) = 3 > 0 h ( 2 ) = 3 > 0 and h ( 5 ) = โˆ’3 < 0 . h ( 5 ) = โˆ’3 < 0 . Therefore, by the IVT, there is a value x = c x = c such that h ( c ) = 0 . h ( c ) = 0 .

The function f ( x ) = 2 x โˆ’ x 3 f ( x ) = 2 x โˆ’ x 3 is continuous over the interval [ 1.25 , 1.375 ] [ 1.25 , 1.375 ] and has opposite signs at the endpoints.

b. It is not possible to redefine f ( 1 ) f ( 1 ) since the discontinuity is a jump discontinuity.

Answers may vary; see the following example:

False. It is continuous over ( โˆ’ โˆž , 0 ) โˆช ( 0 , โˆž ) . ( โˆ’ โˆž , 0 ) โˆช ( 0 , โˆž ) .

False. Consider f ( x ) = { x if x โ‰  0 4 if x = 0 . f ( x ) = { x if x โ‰  0 4 if x = 0 .

False. IVT only says that there is at least one solution; it does not guarantee that there is exactly one. Consider f ( x ) = cos ( x ) f ( x ) = cos ( x ) on [ โˆ’ ฯ€ , 2 ฯ€ ] . [ โˆ’ ฯ€ , 2 ฯ€ ] .

False. The IVT does not work in reverse! Consider ( x โˆ’ 1 ) 2 ( x โˆ’ 1 ) 2 over the interval [ โˆ’2 , 2 ] . [ โˆ’2 , 2 ] .

R = 0.0001519 m R = 0.0001519 m

D = 345,826 km D = 345,826 km

For all values of a , f ( a ) a , f ( a ) is defined, lim ฮธ โ†’ a f ( ฮธ ) lim ฮธ โ†’ a f ( ฮธ ) exists, and lim ฮธ โ†’ a f ( ฮธ ) = f ( a ) . lim ฮธ โ†’ a f ( ฮธ ) = f ( a ) . Therefore, f ( ฮธ ) f ( ฮธ ) is continuous everywhere.

Section 2.5 Exercises

For every ฮต > 0 , ฮต > 0 , there exists a ฮด > 0 , ฮด > 0 , so that if 0 < | t โˆ’ b | < ฮด , 0 < | t โˆ’ b | < ฮด , then | g ( t ) โˆ’ M | < ฮต | g ( t ) โˆ’ M | < ฮต

For every ฮต > 0 , ฮต > 0 , there exists a ฮด > 0 , ฮด > 0 , so that if 0 < | x โˆ’ a | < ฮด , 0 < | x โˆ’ a | < ฮด , then | ฯ† ( x ) โˆ’ A | < ฮต | ฯ† ( x ) โˆ’ A | < ฮต

ฮด โ‰ค 0.25 ฮด โ‰ค 0.25

ฮด โ‰ค 2 ฮด โ‰ค 2

ฮด โ‰ค 1 ฮด โ‰ค 1

ฮด < 0.3900 ฮด < 0.3900

Let ฮด = ฮต . ฮด = ฮต . If 0 < | x โˆ’ 3 | < ฮต , 0 < | x โˆ’ 3 | < ฮต , then | x + 3 โˆ’ 6 | = | x โˆ’ 3 | < ฮต . | x + 3 โˆ’ 6 | = | x โˆ’ 3 | < ฮต .

Let ฮด = ฮต 4 . ฮด = ฮต 4 . If 0 < | x | < ฮต 4 , 0 < | x | < ฮต 4 , then | x 4 | = x 4 < ฮต . | x 4 | = x 4 < ฮต .

Let ฮด = ฮต 2 . ฮด = ฮต 2 . If 5 โˆ’ ฮต 2 < x < 5 , 5 โˆ’ ฮต 2 < x < 5 , then | 5 โˆ’ x | = 5 โˆ’ x < ฮต . | 5 โˆ’ x | = 5 โˆ’ x < ฮต .

Let ฮด = ฮต / 5 . ฮด = ฮต / 5 . If 1 โˆ’ ฮต / 5 < x < 1 , 1 โˆ’ ฮต / 5 < x < 1 , then | f ( x ) โˆ’ 3 | = 5 x โˆ’ 5 < ฮต . | f ( x ) โˆ’ 3 | = 5 x โˆ’ 5 < ฮต .

Let ฮด = 3 M . ฮด = 3 M . If 0 < | x + 1 | < 3 M , 0 < | x + 1 | < 3 M , then f ( x ) = 3 ( x + 1 ) 2 > M . f ( x ) = 3 ( x + 1 ) 2 > M .

0.328 cm, ฮต = 8 , ฮด = 0.33 , a = 12 , L = 144 ฮต = 8 , ฮด = 0.33 , a = 12 , L = 144

lim x โ†’ a f x + lim x โ†’ a g x = L + M lim x โ†’ a f x + lim x โ†’ a g x = L + M

Review Exercises

False. A removable discontinuity is possible.

8 / 7 8 / 7

2 / 3 2 / 3

Since โˆ’1 โ‰ค cos ( 2 ฯ€ x ) โ‰ค 1 , โˆ’1 โ‰ค cos ( 2 ฯ€ x ) โ‰ค 1 , then โˆ’ x 2 โ‰ค x 2 cos ( 2 ฯ€ x ) โ‰ค x 2 . โˆ’ x 2 โ‰ค x 2 cos ( 2 ฯ€ x ) โ‰ค x 2 . Since lim x โ†’ 0 x 2 = 0 = lim x โ†’ 0 โˆ’ x 2 , lim x โ†’ 0 x 2 = 0 = lim x โ†’ 0 โˆ’ x 2 , it follows that lim x โ†’ 0 x 2 cos ( 2 ฯ€ x ) = 0 . lim x โ†’ 0 x 2 cos ( 2 ฯ€ x ) = 0 .

[ 2 , โˆž ] [ 2 , โˆž ]

c = โˆ’1 c = โˆ’1

ฮด = ฮต 3 ฮด = ฮต 3

0 m / sec 0 m / sec

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  • Authors: Gilbert Strang, Edwin โ€œJedโ€ Herman
  • Publisher/website: OpenStax
  • Book title: Calculus Volume 1
  • Publication date: Mar 30, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/calculus-volume-1/pages/1-introduction
  • Section URL: https://openstax.org/books/calculus-volume-1/pages/chapter-2

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