Checkpoint
2.25
12.006001
17 unit2
lim x → 1 1 x − 1 x − 1 = −1 lim x → 1 1 x − 1 x − 1 = −1
lim x → 2 h ( x ) = −1 . lim x → 2 h ( x ) = −1 .
limx→2|x2−4|x−2limx→2|x2−4|x−2 does not exist.
a. limx→2−|x2−4|x−2=−4;limx→2−|x2−4|x−2=−4; b. limx→2+|x2−4|x−2=4limx→2+|x2−4|x−2=4
a. limx→0−1x2=+∞;limx→0−1x2=+∞; b. limx→0+1x2=+∞;limx→0+1x2=+∞; c. limx→01x2=+∞limx→01x2=+∞
a. limx→2−1(x−2)3=−∞;limx→2−1(x−2)3=−∞; b. limx→2+1(x−2)3=+∞;limx→2+1(x−2)3=+∞; c. limx→21(x−2)3limx→21(x−2)3 DNE. The line x=2x=2 is the vertical asymptote of f(x)=1/(x−2)3.f(x)=1/(x−2)3.
Does not exist.
11 10 11 10
−13;
1 3 1 3
1 4 1 4
−1;
1 4 1 4
limx→−1−f(x)=−1limx→−1−f(x)=−1
+∞
0
0
f is not continuous at 1 because f(1)=2≠3=limx→1f(x).f(1)=2≠3=limx→1f(x).
f(x)f(x) is continuous at every real number.
Discontinuous at 1; removable
[ −3 , + ∞ ) [ −3 , + ∞ )
0
f(0)=1>0,f(1)=−2<0;f(x)f(0)=1>0,f(1)=−2<0;f(x) is continuous over [0,1].[0,1]. It must have a zero on this interval.
Let ε>0;ε>0; choose δ=ε3;δ=ε3; assume 0<|x−2|<δ.0<|x−2|<δ.
Thus, |(3x−2)−4|=|3x−6|=|3|·|x−2|<3·δ=3·(ε/3)=ε.|(3x−2)−4|=|3x−6|=|3|·|x−2|<3·δ=3·(ε/3)=ε.
Therefore, limx→23x−2=4.limx→23x−2=4.
Choose δ=min{9−(3−ε)2,(3+ε)2−9}.δ=min{9−(3−ε)2,(3+ε)2−9}.
| x 2 − 1 | = | x − 1 | · | x + 1 | < ε / 3 · 3 = ε | x 2 − 1 | = | x − 1 | · | x + 1 | < ε / 3 · 3 = ε
δ = ε 2 δ = ε 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
3
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
π
a. −0.95238095; b. −0.99009901; c. −0.99502488; d. −0.99900100; e. (−1;.0500000,−0;.95238095); f. (−1;.0100000,−0;.9909901); g. (−1;.0050000,−0;.99502488); h. (1.0010000,−0;.99900100); i. −0.95238095; j. −0.99009901; k. −0.99502488; l. −0.99900100
y = − x − 2 y = − x − 2
−49 m/sec (velocity of the ball is 49 m/sec downward)
5.2 m/sec
−9.8 m/sec
6 m/sec
Under, 1 unit2; over: 4 unit2. The exact area of the two triangles is 12(1)(1)+12(2)(2)=2.5units2.12(1)(1)+12(2)(2)=2.5units2.
Under, 0.96 unit2; over, 1.92 unit2. The exact area of the semicircle with radius 1 is π(1)22=π2π(1)22=π2 unit2.
Approximately 1.3333333 unit2
Section 2.2 Exercises
limx→1f(x)limx→1f(x) does not exist because limx→1−f(x)=−2≠limx→1+f(x)=2.limx→1−f(x)=−2≠limx→1+f(x)=2.
lim x → 0 ( 1 + x ) 1 / x = 2.7183 lim x → 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; limx→0sin2xx=2limx→0sin2xx=2
lim x → 0 sin a x x = a lim x → 0 sin a x x = a
a. −0.80000000; b. −0.98000000; c. −0.99800000; d. −0.99980000; e. −1.2000000; f. −1.0200000; g. −1.0020000; h. −1.0002000; limx→1(1−2x)=−1limx→1(1−2x)=−1
a. −37.931934; b. −3377.9264; c. −333,777.93; d. −33,337,778; e. −29.032258; f. −3289.0365; g. −332,889.04; h. −33,328,889 limx→0z−1z2(z+3)=−∞limx→0z−1z2(z+3)=−∞
a. 0.13495277; b. 0.12594300; c. 0.12509381; d. 0.12500938; e. 0.11614402; f. 0.12406794; g. 0.12490631; h. 0.12499063; ∴limx→21−2xx2−4=0.1250=18∴limx→21−2xx2−4=0.1250=18
a. 10.00000; b. 100.00000; c. 1000.0000; d. 10,000.000; Guess: limα→0+1αcos(πα)=∞,limα→0+1αcos(πα)=∞, actual: DNE
False; limx→−2+f(x)=+∞limx→−2+f(x)=+∞
False; limx→6f(x)limx→6f(x) DNE since limx→6−f(x)=2limx→6−f(x)=2 and limx→6+f(x)=5.limx→6+f(x)=5.
2
1
1
DNE
0
DNE
2
3
DNE
0
−2
DNE
0
Answers may vary.
Answers may vary.
a. ρ2ρ2 b. ρ1ρ1 c. DNE unless ρ1=ρ2.ρ1=ρ2. As you approach xSFxSF from the left, you are in the high-density area of the shock. When you approach from the right, you have not experienced the “shock” yet and are at a lower density.
Section 2.3 Exercises
Use constant multiple law and difference law: limx→0(4x2−2x+3)=4limx→0x2−2limx→0x+limx→03=3limx→0(4x2−2x+3)=4limx→0x2−2limx→0x+limx→03=3
Use root law: limx→−2x2−6x+3=limx→−2(x2−6x+3)=19limx→−2x2−6x+3=limx→−2(x2−6x+3)=19
49
1
− 5 7 − 5 7
limx→4x2−16x−4=16−164−4=00;limx→4x2−16x−4=16−164−4=00; then, limx→4x2−16x−4=limx→4(x+4)(x−4)x−4=8limx→4x2−16x−4=limx→4(x+4)(x−4)x−4=8
limx→63x−182x−12=18−1812−12=00;limx→63x−182x−12=18−1812−12=00; then, limx→63x−182x−12=limx→63(x−6)2(x−6)=32limx→63x−182x−12=limx→63(x−6)2(x−6)=32
limx→9t−9t−3=9−93−3=00;limx→9t−9t−3=9−93−3=00; then, limt→9t−9t−3=limt→9t−9t−3t+3t+3=limt→9(t+3)=6limt→9t−9t−3=limt→9t−9t−3t+3t+3=limt→9(t+3)=6
limθ→πsinθtanθ=sinπtanπ=00;limθ→πsinθtanθ=sinπtanπ=00; then, limθ→πsinθtanθ=limθ→πsinθsinθcosθ=limθ→πcosθ=−1limθ→πsinθtanθ=limθ→πsinθsinθcosθ=limθ→πcosθ=−1
limx→1/22x2+3x−22x−1=12+32−21−1=00;limx→1/22x2+3x−22x−1=12+32−21−1=00; then, limx→1/22x2+3x−22x−1=limx→1/2(2x−1)(x+2)2x−1=52limx→1/22x2+3x−22x−1=limx→1/2(2x−1)(x+2)2x−1=52
−∞
−∞
lim x → 6 2 f ( x ) g ( x ) = 2 lim x → 6 f ( x ) lim x → 6 g ( x ) = 72 lim x → 6 2 f ( x ) g ( x ) = 2 lim x → 6 f ( x ) lim x → 6 g ( x ) = 72
lim x → 6 ( f ( x ) + 1 3 g ( x ) ) = lim x → 6 f ( x ) + 1 3 lim x → 6 g ( x ) = 7 lim x → 6 ( f ( x ) + 1 3 g ( x ) ) = lim x → 6 f ( x ) + 1 3 lim x → 6 g ( x ) = 7
lim x → 6 g ( x ) − f ( x ) = lim x → 6 g ( x ) − lim x → 6 f ( x ) = 5 lim x → 6 g ( x ) − f ( x ) = lim x → 6 g ( x ) − lim x → 6 f ( x ) = 5
lim x → 6 [ ( x + 1 ) f ( x ) ] = ( lim x → 6 ( x + 1 ) ) ( lim x → 6 f ( x ) ) = 28 lim x → 6 [ ( x + 1 ) f ( x ) ] = ( lim x → 6 ( x + 1 ) ) ( lim x → 6 f ( x ) ) = 28
a. 9; b. 7
a. 1; b. 1
lim x → −3 − ( f ( x ) − 3 g ( x ) ) = lim x → −3 − f ( x ) − 3 lim x → −3 − g ( x ) = 0 + 6 = 6 lim x → −3 − ( f ( x ) − 3 g ( x ) ) = lim x → −3 − f ( x ) − 3 lim x → −3 − g ( x ) = 0 + 6 = 6
lim x → −5 2 + g ( x ) f ( x ) = 2 + ( lim x → −5 g ( x ) ) lim x → −5 f ( x ) = 2 + 0 2 = 1 lim x → −5 2 + g ( x ) f ( x ) = 2 + ( lim x → −5 g ( x ) ) lim x → −5 f ( x ) = 2 + 0 2 = 1
lim x → 1 f ( x ) − g ( x ) 3 = lim x → 1 f ( x ) − lim x → 1 g ( x ) 3 = 2 + 5 3 = 7 3 lim x → 1 f ( x ) − g ( x ) 3 = lim x → 1 f ( x ) − lim x → 1 g ( x ) 3 = 2 + 5 3 = 7 3
lim x → −9 ( x f ( x ) + 2 g ( x ) ) = ( lim x → −9 x ) ( lim x → −9 f ( x ) ) + 2 lim x → −9 ( g ( x ) ) = ( −9 ) ( 6 ) + 2 ( 4 ) = −46 lim x → −9 ( x f ( x ) + 2 g ( x ) ) = ( lim x → −9 x ) ( lim x → −9 f ( x ) ) + 2 lim x → −9 ( g ( x ) ) = ( −9 ) ( 6 ) + 2 ( 4 ) = −46
The limit is zero.
a.
∞. ਕਣ q ਦੇ ਨੇੜੇ ਪਹੁੰਚਣ 'ਤੇ ਬਿਜਲਈ ਖੇਤਰ ਦੀ ਮਾਤਰਾ ਅਨੰਤ ਹੋ ਜਾਂਦੀ ਹੈ। ਰਿਣਾਤਮਕ ਦੂਰੀ ਦਾ ਮੁਲਾਂਕਣ ਕਰਨਾ ਭੌਤਿਕ ਤੌਰ 'ਤੇ ਕੋਈ ਅਰਥ ਨਹੀਂ ਰੱਖਦਾ।
2.4 ਭਾਗ ਅਭਿਆਸ
ਇਹ ਫੰਕਸ਼ਨ ਅੰਤਰਾਲ (0,∞) ਵਿੱਚ ਸਾਰੇ x ਲਈ ਪਰਿਭਾਸ਼ਿਤ ਹੈ।
x=0 'ਤੇ ਹਟਾਉਣਯੋਗ ਅਸੰਤਤਤਾ; x=1 'ਤੇ ਅਨੰਤ ਅਸੰਤਤਤਾ
x=ln2 'ਤੇ ਅਨੰਤ ਅਸੰਤਤਤਾ
x=(2k+1)π/4, k=0,±1,±2,±3,… ਲਈ ਅਨੰਤ ਅਸੰਤਤਤਾਵਾਂ
ਨਹੀਂ। ਇਹ ਇੱਕ ਹਟਾਉਣਯੋਗ ਅਸੰਤਤਤਾ ਹੈ।
ਹਾਂ। ਇਹ ਨਿਰੰਤਰ ਹੈ।
ਹਾਂ। ਇਹ ਨਿਰੰਤਰ ਹੈ।
k = −5
k = −1
k = 16/3
ਕਿਉਂਕਿ s ਅਤੇ y=t ਦੋਵੇਂ ਹਰ ਥਾਂ ਨਿਰੰਤਰ ਹਨ, ਇਸ ਲਈ h(t)=s(t)−t ਹਰ ਥਾਂ ਨਿਰੰਤਰ ਹੈ ਅਤੇ, ਖਾਸ ਤੌਰ 'ਤੇ, ਇਹ ਬੰਦ ਅੰਤਰਾਲ [2,5] 'ਤੇ ਨਿਰੰਤਰ ਹੈ। ਨਾਲ ਹੀ, h(2)=3>0 ਅਤੇ h(5)=−3<0। ਇਸ ਲਈ, IVT ਦੁਆਰਾ, ਇੱਕ ਮੁੱਲ x=c ਹੈ ਜਿਸ ਲਈ h(c)=0।
ਫੰਕਸ਼ਨ f(x)=2x−x³ ਅੰਤਰਾਲ [1.25,1.375] 'ਤੇ ਨਿਰੰਤਰ ਹੈ ਅਤੇ ਸਿਰਿਆਂ 'ਤੇ ਵਿਰੋਧੀ ਚਿੰਨ੍ਹ ਰੱਖਦਾ ਹੈ।
a.
b. f(1) ਨੂੰ ਮੁੜ ਪਰਿਭਾਸ਼ਿਤ ਕਰਨਾ ਸੰਭਵ ਨਹੀਂ ਹੈ ਕਿਉਂਕਿ ਅਸੰਤਤਤਾ ਇੱਕ ਛਾਲ ਅਸੰਤਤਤਾ ਹੈ।
ਜਵਾਬ ਵੱਖ-ਵੱਖ ਹੋ ਸਕਦੇ ਹਨ; ਹੇਠਾਂ ਦਿੱਤੇ ਉਦਾਹਰਨ ਨੂੰ ਦੇਖੋ:
ਜਵਾਬ ਵੱਖ-ਵੱਖ ਹੋ ਸਕਦੇ ਹਨ; ਹੇਠਾਂ ਦਿੱਤੇ ਉਦਾਹਰਨ ਨੂੰ ਦੇਖੋ:
ਗਲਤ। ਇਹ (−∞,0)∪(0,∞) 'ਤੇ ਨਿਰੰਤਰ ਹੈ।
ਗਲਤ। f(x)={x ਜੇ x≠0; 4 ਜੇ x=0 'ਤੇ ਵਿਚਾਰ ਕਰੋ।
ਗਲਤ। IVT ਸਿਰਫ ਇਹ ਕਹਿੰਦਾ ਹੈ ਕਿ ਘੱਟੋ-ਘੱਟ ਇੱਕ ਹੱਲ ਹੈ; ਇਹ ਗਾਰੰਟੀ ਨਹੀਂ ਦਿੰਦਾ ਕਿ ਸਿਰਫ ਇੱਕ ਹੀ ਹੈ। [−π,2π] 'ਤੇ f(x)=cos(x) 'ਤੇ ਵਿਚਾਰ ਕਰੋ।
ਗਲਤ। IVT ਉਲਟਾ ਕੰਮ ਨਹੀਂ ਕਰਦਾ! [−2,2] 'ਤੇ (x−1)² 'ਤੇ ਵਿਚਾਰ ਕਰੋ।
R = 0.0001519 ਮੀ
D = 345,826 km D = 345,826 km
For all values of a,f(a)a,f(a) is defined, limθ→af(θ)limθ→af(θ) exists, and limθ→af(θ)=f(a).limθ→af(θ)=f(a). Therefore, f(θ)f(θ) is continuous everywhere.
Nowhere
Section 2.5 Exercises
For every ε>0,ε>0, there exists a δ>0,δ>0, so that if 0<|t−b|<δ,0<|t−b|<δ, then |g(t)−M|<ε|g(t)−M|<ε
For every ε>0,ε>0, there exists a δ>0,δ>0, so that if 0<|x−a|<δ,0<|x−a|<δ, then |φ(x)−A|<ε|φ(x)−A|<ε
δ ≤ 0.25 δ ≤ 0.25
δ ≤ 2 δ ≤ 2
δ ≤ 1 δ ≤ 1
δ < 0.3900 δ < 0.3900
Let δ=ε.δ=ε. If 0<|x−3|<ε,0<|x−3|<ε, then |x+3−6|=|x−3|<ε.|x+3−6|=|x−3|<ε.
Let δ=ε4.δ=ε4. If 0<|x|<ε4,0<|x|<ε4, then |x4|=x4<ε.|x4|=x4<ε.
Let δ=ε2.δ=ε2. If 5−ε2<x<5,5−ε2<x<5, then |5−x|=5−x<ε.|5−x|=5−x<ε.
Let δ=ε/5.δ=ε/5. If 1−ε/5<x<1,1−ε/5<x<1, then |f(x)−3|=5x−5<ε.|f(x)−3|=5x−5<ε.
Let δ=3M.δ=3M. If 0<|x+1|<3M,0<|x+1|<3M, then f(x)=3(x+1)2>M.f(x)=3(x+1)2>M.
The engineer must cut within 0.328 cm of 12 cm on each side; ε=8,δ=0.328,a=12,L=144ε=8,δ=0.328,a=12,L=144
Answers may vary.
0
lim x → a f x + lim x → a g x = L + M lim x → a f x + lim x → a g x = L + M
Answers may vary.
Review Exercises
False
False. A removable discontinuity is possible.
5
੧. ੮ / ੭ ੮ / ੭
DNE
੨. ੨ / ੩ ੨ / ੩
੩. −੪;
੪. ਕਿਉਂਕਿ −੧≤cos(੨πx)≤੧, ਇਸ ਲਈ −x²≤x²cos(੨πx)≤x². ਕਿਉਂਕਿ limx→੦x²=੦=limx→੦−x², ਇਸ ਲਈ limx→੦x²cos(੨πx)=੦.
੫. [ ੨ , ∞ ]
੬. c = −੧
੭. δ = ε / ੩
੮. ੦ ਮੀਟਰ / ਸੈਕਿੰਡ