Homework #9 Review Packet for Exam #1

HOMEWORK #9 Review Packet for Exam #1 Due Wednesday October 2 at the beginning of class. Limit Practice Problems Evaluate the following limits. Be clear if the limit equals a finite value, Does Not Exist, or is +∞ or −∞. Always justify your work: 16 = DOES NOT EXIST since RHL 6= LHL w→0 w 16 16 RHL: lim = + = +∞ + 0 w→0 w 16 16 LHL: lim = − = −∞ 0 w→0− w 1. lim 3−t = DOES NOT EXIST since RHL 6= LHL t→2 t − 2 3−2 1 3−t = + = + = +∞ RHL: lim 0 0 t→2+ t − 2 3−2 1 3−t LHL: lim = − = − = −∞ 0 0 t→2− t − 2 2. lim 3−t = +∞ since RHL = LHL t→2 (t − 2)2 3−2 1 3−t = + 2 = + = +∞ RHL: lim 2 (0 ) 0 t→2+ (t − 2) 3−t 3−2 1 LHL: lim = − 2 = + = +∞ 2 − (0 ) 0 t→2 (t − 2) 3. lim (x + 2)2 = DOES NOT EXIST since RHL 6= LHL x→4 x2 − 3x − 4 (x + 2)2 (4 + 2)2 36 (x + 2)2 = lim = + = + = +∞ RHL: lim 2 0 (4 + 1) 0 (5) x→4+ (x − 4)(x + 1) x→4+ x − 3x − 4 4. lim LHL: lim x→4− (x + 2)2 (4 + 2)2 36 (x + 2)2 = lim = = − = −∞ x2 − 3x − 4 x→4− (x − 4)(x + 1) 0− (4 + 1) 0 (5) 0 0 1 DSP 1 x−4 x−4 = lim = 5. lim 2 = lim x→4 x + 1 x→4 x − 3x − 4 x→4 (x − 4)(x + 1) 5 0 (x − 4)(x + 2) x2 − 2x − 8 0 6 x + 2 DSP 4 + 2 = lim 6. lim 2 = lim = = x→4 (x − 4)(x + 1) x→4 x − 3x − 4 x→4 x + 1 4+1 5 x2 − 4x − 12 3 1 − 4 − 12 DSP −15 = = = 2 x→1 x − 3x − 18 1 − 3 − 18 −20 4 7. lim x2 − 4x − 12 DSP −12 2 = = x→0 x2 − 3x − 18 −18 3 8. lim 1 x+2 = DOES NOT EXIST since RHL 6= LHL x+3 x+2 −1 RHL: lim = + = −∞ + 0 x→−3 x + 3 x+2 −1 LHL: lim = − = +∞ 0 x→−3− x + 3 9. lim x→−3 x2 − 4x − 12 DSP 4 + 8 − 12 0 = = = 0 2 x→−2 x − 3x − 18 4 + 6 − 18 −8 10. lim 0 4 x − 4 DSP −4 x(x − 4) x2 − 4x 0 = lim = = = lim 11. lim 2 x→0 x − 7 x→0 x(x − 7) x→0 x − 7x −7 7 x2 − 9 = DOES NOT EXIST since RHL 6= LHL x→3 |x − 3| 12. lim RHL: lim x→3+ x2 − 9 (x − 3)(x + 3) DSP = lim = lim x + 3 = 6 |x − 3| x→3+ x−3 x→3+ x2 − 9 (x − 3)(x + 3) DSP = lim = lim −(x + 3) = −6 −(x − 3) x→3− |x − 3| x→3− x→3−   if x − 3 ≥ 0 if x ≥ 3  x−3  x−3 Recall: |x − 3| = =   −(x − 3) if x − 3 < 0 −(x − 3) if x < 3 LHL: lim x2 + 6x + 5 = DOES NOT EXIST since RHL 6= LHL x→−5 |x + 5| 13. lim RHL: lim x→(−5)+ x2 + 6x + 5 (x + 5)(x + 1) DSP = lim = lim x + 1 = −4 + + |x + 5| x+5 x→(−5) x→(−5) (x + 5)(x + 1) x2 + 6x + 5 DSP = lim = lim −(x + 1) = 4 |x + 5| −(x + 5) x→(−5)− x→(−5)− x→(−5)−   if x + 5 ≥ 0 if x ≥ −5  x+5  x+5 Recall: |x + 5| = =   −(x + 5) if x + 5 < 0 −(x + 5) if x < −5 LHL: lim 0 0 t + 1 DSP 1 + 1 t2 − 1 (t − 1)(t + 1) 2 = lim = = − 14. lim 2 = lim t→1 t − 10 t→1 t − 11t + 10 t→1 (t − 10)(t − 1) 1 − 10 9 t2 1 DSP = 1 = t→1 t2 + t − 1 1+1−1 15. lim 0 2009(t2 + 6t + 5) 0 2009(t + 5) DSP 2009(4) 2009(t + 5)(t + 1) 16. lim = = lim = −8036 = lim 2 t→−1 t→−1 t→−1 t +t t(t + 1) t −1 0 (x − 9)(x − 1) x2 − 10x + 9 0 8 x − 1 DSP 9 − 1 = lim 17. lim 2 = lim = = x→9 (x + 10)(x − 9) x→9 x + x − 90 x→9 x + 10 9 + 10 19 DSP 18. lim t500 + t400 + t300 + t200 + t100 = 5 t→1 2 0 x2 − x − 6 0 5 x + 2 DSP 3 + 2 (x − 3)(x + 2) 19. lim 2 = lim = = = lim x→3 x − 2x − 3 x→3 x + 1 x→3 (x − 3)(x + 1) 3+1 4 √ √ 0 x−1 x−1 0 x+3+2 (x − 1)( x + 3 + 2) = lim √ 20. lim √ ·√ = lim x→1 x→1 (x + 3) − 4 x+3−2 x+3−2 x + 3 + 2 x→1 √ √ (x − 1)( x + 3 + 2) L.L. = lim x + 3 + 2 = 4 = lim x→1 x→1 x−1 0 √ √ 9x − x2 0 x(9 − x)(3 + x) 9x − x2 3 + x √ √ · √ = lim 21. lim = lim x→9 3 − x x→9 x→9 3 − x 3 + x 9−x √ L.L. = lim x(3 + x) = 9(3 + 3) = 54 x→9 22. lim x→−1 23. lim x→5 5 DSP 5 = 1−x 2 6x = 5−x DOES NOT EXIST since RHL 6= LHL 6x 30 = − = −∞ 5−x 0 30 6x = + = +∞ LHL: lim − 0 x→5 5 − x RHL: lim x→5+ 0 x−7 x2 − 9x + 14 0 (x − 2)(x − 7) = lim 24. lim 2 = lim x→2 x − 2 x→2 x − 4x + 4 x→2 (x − 2)(x − 2) −5 x−7 = lim + = −∞ RHL: lim x→2+ 0 x→2+ x − 2 x−7 x−7 −5 LHL: lim = lim = lim − = +∞ x→2− x − 2 x→2− x − 2 x→2− 0 x2 − 4 = x→2 |x − 2| 25. lim RHL: lim x→2+ DNE b/c RHL6= LHL DNE b/c RHL 6= LHL . x2 − 4 x2 − 4 (x − 2)(x + 2) DSP = lim = lim = lim x + 2 = 4 |x − 2| x→2+ x − 2 x−2 x→2+ x→2+ x2 − 4 x2 − 4 (x − 2)(x + 2) DSP = lim = lim = lim −(x + 2) = −4 −(x − 2) x→2− |x − 2| x→2− −(x − 2) x→2− x→2−   if x ≥ 2 if x − 2 ≥ 0  x−2  x−2 = Recall: |x − 2| =   −(x − 2) if x < 2 −(x − 2) if x − 2 < 0 LHL: lim √ 0 √ x+6−30 x+6−3 x+6−9 √ 26. lim 2 = lim 2 = lim x→3 x − x − 6 x→3 x − x − 6 x→3 (x − 3)(x + 2)( x + 6 + 3) = lim x→3 1 1 x−3 1 L.L. √ √ = = = lim x→3 5(3 + 3) 30 (x − 3)(x + 2)( x + 6 + 3) (x + 2)( x + 6 + 3) 3 1 1 − 7 x 27. lim x→7 x − 7 0 0 7 x−7 x − 1 x−7 x−7 7x 7x = lim 7x = lim · = lim = lim x→7 x − 7 x→7 7x x→7 x − 7 x − 7 x→7 (7x)(x − 7) 1 DSP 1 = x→7 7x 49 = lim 1 1 − 2 − x 8 28. lim x→−6 x+6 0 0 6+x 8 − (2 − x) (2 − x)(8) (2 − x)(8) = lim = lim x→−6 x→−6 x+6 x+6 1 6+x DSP 1 = lim = x→−6 (2 − x)(8) x→−6 (2 − x)(8)(x + 6) 64 = lim √ 0 √ √  x+1−20 x+1−2 x+1+2 x+1−4 √ 29. lim = lim · √ = lim x→3 x→3 x→3 (3 − x)( x + 1 + 2) 3−x 3−x x+1+2 x−3 x−3 √ √ = lim = lim x→3 (3 − x)( x + 1 + 2) x→3 −(x − 3)( x + 1 + 2) = lim √ x→3 1 −1 L.L. −1 = = − 2 + 2 4 x+1+2 0 √ √   2+ x−3 x2 − 49 (x2 − 49)(2 + x − 3) x2 − 49 0 √ √ √ · = lim = lim 30. lim x→7 2 − x − 3 x→7 x→7 2 − x − 3 4 − (x − 3) 2+ x−3 √ √ (x − 7)(x + 7)(2 + x − 3) (x − 7)(x + 7)(2 + x − 3) = lim = lim x→7 x→7 7−x −(x − 7) √ L.L. = lim −(x + 7)(2 + x − 3) = −(14)(2 + 2) = −56 x→7 1 1 √ − x + 20 5 31. lim x→5 x−5 0 0 √ 5 − x + 20 √ √ √   5 − x + 20 5 + x + 20 5( x + 20) √ = lim √ · = lim x→5 5( x + 20)(x − 5) x→5 x−5 5 + x + 20 25 − (x + 20) 5−x √ √ = lim √ x→5 5( x + 20)(x − 5)(5 + x + 20) x→5 5( x + 20)(x − 5)(5 + x + 20) = lim √ −(x − 5) √ x→5 5( x + 20)(x − 5)(5 + x + 20) = lim √ −1 −1 −1 L.L. √ = = x→5 5( x + 20)(5 + x + 20) 5(5)(5 + 5) 250 Challenge! = lim √ Functions and Limit Practice Problems Evaluate the following limits: x−1 = x→1 g(x2 ) − 3 32. Let g(x) = 2x + 1. Compute lim 4 lim x→1 x−1 x−1 x−1 1 DSP 1 = = lim = lim = lim (2x2 + 1) − 3 x→1 2x2 − 2 x→1 2(x − 1)(x + 1) x→1 2(x + 1) 4 u2 − 2u u2 − 2u = lim u→2 G(u − 3) u→2 (u − 3)2 + (u − 3) 33. Let G(u) = u2 + u. Compute lim u2 − 2u u(u − 2) u(u − 2) = lim 2 = lim 2 u→2 u − 6u + 9 + u − 3 u→2 u − 5u + 6 u→2 (u − 3)(u − 2) u DSP 2 = −2 = = lim u→2 u − 3 −1 = lim 34. Let h(y) = y 2 − 3. Compute lim x→−2 x+2 = h(2x) − h(x + 6) x+2 x+2 = lim 2 2 − 3) − ((x + 6) − 3) x→−2 (4x − 3) − (x2 + 12x + 36 − 3) x+2 x+2 x+2 = lim = lim = lim x→−2 3x2 − 12x − 36 x→−2 3(x2 − 4x − 12) x→−2 4x2 − 3 − x2 − 12x − 33 lim x→−2 ((2x)2 1 x+2 1 DSP = − = lim x→−2 3(x − 6) x→−2 3(x − 6)(x + 2) 24 = lim 1 f (t − 3) − 4f (t) 35. Let f (t) = . Compute lim = t→4 t t−4       4 1 t − 4(t − 3) −3t + 12 − t−3 t (t − 3)t (t − 3)t lim = lim = lim t→4 t→4 t→4 t−4 t−4 t−4     −3t + 12 1 −3(t − 4) = lim · = lim t→4 t→4 (t − 3)t(t − 4) (t − 3)t t−4 = lim t→4 3 −3 DSP −3 = − = (t − 3)t (1)(4) 4 More Functions 36. Let f (x) = following: √ x, g(x) = x2 + 4, and h(x) = 1 . Compute (and simplify, if possible) the x √
(a) f ◦ g(x) = f (g(x)) = f x2 + 4 = x2 + 4 √ √ (b) g ◦ f (x) = g(f (x)) = g ( x) = ( x)2 + 4 = x + 4 √ (c) h ◦ g ◦ f (x) = h(g(f (x)) = h(g( x)) = h(x + 4) = 1 x+4 (d) g ◦ g(x) = g(g(x)) = g(x2 + 4) = (x2 + 4)2 + 4 = x4 + 8x2 + 16 + 4 = x4 + 8x2 + 20 5 Piece-wise defined functions Consider each of the following piecewise defined functions. Answer the related questions. Justify your answers please.  if x < 0  x x2 if 0 ≤ x ≤ 2 37. Let f (x) =  8 − x if x > 2 Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (x) = DNE b/c LHL6= LHL x→2 RHL: lim f (x) = 6 x→2+ LHL: lim f (x) = 4 x→2− lim f (x) = 0 x→0 RHL: lim f (x) = 0 x→0+ LHL: lim f (x) = 0 x→0− 6 f is discontinuous at x = 2 since lim f (x) DNE x→2  x−1    1 38. Let f (x) = 3    √ x if if if if x<2 24 Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (x) = −1 b/c RHL=LHL x→0 RHL: lim f (x) = −1 x→0+ LHL: lim f (x) = −1 x→0− lim f (x) = 1 b/c RHL=LHL x→2 RHL: lim f (x) = 1 x→2+ LHL: lim f (x) = 1 x→2− 7 lim f (x) = DNE b/c LHL6= LHL x→4 RHL: lim f (x) = 2 x→4+ LHL: lim f (x) = 1 x→4− f (4) = 3 f is discontinuous at x = 2 because f (2) is undefined f is discontinuous at x = 4 because lim f (x) DNE b/c LHL6= LHL x→4  8       x+2 2 39. Let h(x) = 1   x−4   2   √ x if x < 0 if x = 0 if 0 < x ≤ 16 if x > 16 Sketch the graph. Find the numbers at which h is discontinuous. Evaluate: 8 lim h(x) = DNE b/c RHL6= LHL x→−2 RHL: LHL: lim h(x) = +∞ x→−2+ lim h(x) = −∞ x→−2− lim h(x) = DNE b/c RHL6= LHL x→0 RHL: lim h(x) = −4 x→0+ LHL: lim h(x) = 4 x→0− lim h(x) = 4 b/c RHL=LHL x→16 RHL: lim h(x) = 4 x→16+ LHL: lim h(x) = 4 x→16− h is discontinuous at x = −2 because lim h(x) DNE or because f (−2) is undefined x→−2 h is discontinuous at x = 0 because lim h(x) DNE x→0 Note h is continuous at x = 16 because lim h(x) = h(16) x→16 1 if x > 4 x−4 40. Let F (x) = 3 − x if 0 < x < 4   3 − x2 if x < 0    Sketch the graph. 9 Find the numbers at which F is discontinuous. Evaluate: lim F (x) = 3 b/c RHL=LHL x→0 RHL: lim F (x) = 3 x→0+ LHL: lim F (x) = 3 x→0− lim F (x) = DNE b/c RHL 6= LHL x→4 RHL: lim F (x) = +∞ x→4+ LHL: lim F (x) = −1 x→4− f is discontinuous at x = 0 because f (0) is undefined f is discontinuous at x = 4 because f (4) is undefined or because lim F (x) DNE x→4 10