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Limits and Continuity: Get Stronger Answer Key

The Limit Laws

In the following exercises (1-2), use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

  1. Use constant multiple law and difference law: limx0(4x22x+3)=4limx0x22limx0x+limx03=3
  2. Use root law: limx2x26x+3=limx2(x26x+3)=19
  3. 49
  4. 1
  5. 57
  6. limx4x216x4=161644=00; then, limx4x216x4=limx4(x+4)(x4)x4=8
  7. limx63x182x12=18181212=00; then, limx63x182x12=limx63(x6)2(x6)=32
  8. limt9t9t3=9933=00; then, limt9t9t3=limt9t9t3t+3t+3=limt9(t+3)=6
  9. limθπsinθtanθ=sinπtanπ=00; then, limθπsinθtanθ=limθπsinθsinθcosθ=limθπcosθ=1.
  10. limx1/22x2+3x22x1=12+32211=00; then, limx1/22x2+3x22x1=limx1/2(2x1)(x+2)2x1=52
  11. limx62f(x)g(x)=2limx6f(x)limx6g(x)=72
  12. limx6(f(x)+13g(x))=limx6f(x)+13limx6g(x)=7
  13. limx6g(x)f(x)=limx6g(x)limx6f(x)=5
  14. limx6[(x+1)f(x)]=(limx6(x+1))(limx6f(x))=28.
  15. The graph of a piecewise function with two segments. The first is the parabola x^2, which exists for x<=3. The vertex is at the origin, it opens upward, and there is a closed circle at the endpoint (3,9). The second segment is the line x+4, which is a linear function existing for x > 3. There is an open circle at (3, 7), and the slope is 1.
    1. 9
    2. 7
  16. The graph of a piecewise function with two segments. The first segment is the parabola x^2 – 2x + 1, for x < 2. It opens upward and has a vertex at (1,0). The second segment is the line 3-x for x>= 2. It has a slope of -1 and an x intercept at (3,0).
    1. 1
    2. 1
  17. limx3(f(x)3g(x))=limx3f(x)3limx3g(x)=0+6=6
  18. limx52+g(x)f(x)=2+(limx5g(x))limx5f(x)=2+02=1
  19. limx13f(x)g(x)=3limx1f(x)limx1g(x)=32+5=37
  20. limx9(xf(x)+2g(x))=(limx9x)(limx9f(x))+2limx9(g(x))=(9)(6)+2(4)=46

Continuity

For the following exercises (1-4), determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

  1. The function is defined for all x in the interval (0,).
  2. Removable discontinuity at x=0; infinite discontinuity at x=1
  3. f(x)=5ex2 Infinite discontinuity at x=ln2
  4. H(x)=tan2x Infinite discontinuities at x=(2k+1)π4, for k=0,±1,±2,±3,
  5. No. It is a removable discontinuity.
  6. Yes. It is continuous.
  7. Yes. It is continuous.
  8. k=5
  9. k=1
  10. k=163
  11. A graph of the given piecewise function containing two segments. The first, x^3, exists for x < 1 and ends with an open circle at (1,1). The second, 3x, exists for x > 1. It beings with an open circle at (1,3).

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

  12. Answers may vary; see the following example:

    A graph of a piecewise function with several segments. The first is an increasing line that exists for x < -8. It ends at an open circle at (-8,-8). The second is an increasing curve that exists from -8 <= x < -6. It begins with a closed circle at (-8, 0 ) and goes to infinity as x goes to -6 from the left. The third is a closed circle at the point (-6, 3). The fourth is a line that exists from -6 < x <= 3. It begins with an open circle at (-6, 2) and ends with a closed circle at (3,2). The fifth is an increasing line starting with an open circle at (3,3). It exists for x > 3.

  13. Answers may vary; see the following example:

    The graph of a piecewise function with two parts. The first part is an increasing curve that exists for x < 1. It ends at (1,1). The second part is an increasing line that exists for x > 1. It begins at (1,3)./li>

  14. False. It is continuous over (,0)(0,).
  15. False. Consider f(x)={x if x04 if x=0
  16. False. Consider f(x)=cos(x) on [π,2π].
  17. False. The IVT does not work in reverse! Consider (x1)2 over the interval [2,2].
  18. For all values of a,f(a) is defined, limθaf(θ) exists, and limθaf(θ)=f(a). Therefore, f(θ) is continuous everywhere.
  19. Nowhere

The Precise Definition of a Limit

In the following exercises (1-2), write the appropriate εδ definition for each of the given statements.

  1. For every ε>0, there exists a δ>0 so that if 0<|tb|<δ, then |g(t)M|<ε
  2. For every ε>0, there exists a δ>0 so that if 0<|xa|<δ, then |ϕ(x)A|<ε
  3. δ0.25
  4. δ2
  5. δ1
  6. δ<0.3900
  7. Let δ=ε. If 0<|x3|<ε, then |x+36|=|x3|<ε.
  8. Let δ=4ε. If 0<|x|<4ε, then |x4|=x4<ε.
  9. Let δ=ε2. If 5ε2<x<5, then |5x|=5x<ε.
  10. Let δ=ε/5. If 1ε/5<x<1, then |f(x)3|=5x5<ε.
  11. Let δ=3N. If 0<|x+1|<3N, then f(x)=3(x+1)2>N.
  12. f(x)g(x)=f(x)+(1)g(x)
  13. Answers may vary.