Limits of Functions at a Point PDF

I H S – Saida 11 Sc. Math. Chapter 10 Limits of functions at a point * Objectives & Prerequisites. * Lesson (Recall, Activities, Text …). ** Solve: … Homework /Practice … * Summary (Notes, Properties, Rules). * Book (Exercises, Problems & Self-evalu...

I H S – Saida 11 Sc. Math. Chapter 10 Limits of functions at a point * Objectives & Prerequisites. * Lesson (Recall, Activities, Text …). ** Solve: … Homework /Practice … * Summary (Notes, Properties, Rules). * Book (Exercises, Problems & Self-evaluation). * WSH (I.H.S exams + …). Adnan Al - Ashkar  11 Sc. Math Chapter 10 Page Adnan Lesson Limits of Functions at a point 2 Ashkar 1 - Definition 𝒍𝒊𝒎𝒇(𝒙) = k (where a R, kR & k: limit of function f at a). 𝒙→𝒂 * If k = ± ∞. Then f(x) has no limit. 1 1 1 1 Eg1: 𝑙𝑖𝑚+ = =+∞ Eg2: 𝑙𝑖𝑚− = =–∞ 𝑥→0 𝑥 0+ 𝑥→0 𝑥 0− Note: 𝒍𝒊𝒎 = 𝒍𝒊𝒎− & 𝒍𝒊𝒎 = 𝒍𝒊𝒎+ 𝒙→ 𝒂 𝒙→𝒂 𝒙→ 𝒂 𝒙→𝒂 𝒙< 𝒂 𝒙> 𝒂 𝟏 Eg3: 𝒍𝒊𝒎 =? 3– 3+ 𝒙→𝟑 𝒙 − 𝟑 x –∞ 3 +∞ x – 3 – + 1 1 1 1 𝑙𝑖𝑚− = = −∞ & 𝑙𝑖𝑚+ = +∞ 𝑥→3 𝑥−3 0− 𝑥→3 𝑥−3 0+ 𝟏 Eg4: 𝒍𝒊𝒎 =? 4– 4+ 𝒙→𝟒 𝟒 − 𝒙 x –∞ 4 +∞ 4 – x + − 1 1 1 1 𝑙𝑖𝑚− = = +∞ & 𝑙𝑖𝑚+ = =−∞ 𝑥→4 4−𝑥 0+ 𝑥→4 4−𝑥 0− 𝟑𝒙−𝟏 Eg5: 𝒍𝒊𝒎 =? 5– 5+ 𝒙→𝟓 𝒙 − 𝟓 x –∞ 5 +∞ x – 5 – + 3𝑥−1 3(5)−1 14 3𝑥 − 1 3(5)−1 14 𝑙𝑖𝑚− = = = −∞ & 𝑙𝑖𝑚+ = = = +∞ 𝑥→5 𝑥−5 0− − 0 𝑥→5 𝑥 −5 0+ 0+ 𝒙𝟐 −𝟓 Eg6: 𝒍𝒊𝒎 =? 1– 1+ 𝒙→𝟏 − 𝒙+𝟏 x –∞ 1 +∞ –x+1 + − 𝑥 2 −5 1−5 −4 𝑥 2 −5 1−5 −4 𝑙𝑖𝑚− = = = −∞ 𝑙𝑖𝑚+ = = = +∞ 𝑥→1 − 𝑥+1 0+ 0+ 𝑥→1 − 𝑥+1 0− 0−  11 Sc. Math Chapter 10 Page Adnan Lesson Limits of Functions at a point 3 Ashkar 𝒙𝟐 +𝟏 𝒙𝟐 +𝟏 Eg7: 𝒍𝒊𝒎 =? & 𝒍𝒊𝒎 =? 𝒙→−𝟐 𝒙𝟐 + 𝒙 − 𝟐 𝒙→𝟏 𝒙𝟐 + 𝒙 − 𝟐 -2– -2+ 1– 1+ x –∞ –2 1 +∞ x2 + x − 2 + – + 𝑥 2 +1 (−2)2 +1 5 𝑙𝑖𝑚 = = = +∞ 𝑥→(−2)− 𝑥2+ 𝑥 −2 0+ 0+ 𝑥 2 +1 (−2)2 +1 5 𝑙𝑖𝑚 = = =−∞ 𝑥→(−2)+ 𝑥 2 + 𝑥 −2 0− 0− 𝑥 2 +1 (1)2 +1 2 𝑙𝑖𝑚− = = = +∞ 𝑥→1 𝑥2+ 𝑥 −2 0+ 0+ 𝑥 2 +1 (1)2 +1 2 𝑙𝑖𝑚− = = = −∞ 𝑥→1 𝑥2+ 𝑥 −2 0− 0− 2 - If 𝒍𝒊𝒎−𝒇(𝒙) = 𝒍𝒊𝒎+𝒇(𝒙) = k → 𝒍𝒊𝒎𝒇(𝒙) = k 𝒙→𝒂 𝒙→𝒂 𝒙→𝒂 If the 𝒍𝒊𝒎𝒇(𝒙) has an indeterminate form, then the indeterminate is removed 𝒙→𝒂 through modifying the expression by factoring (Identities – /using the calculator, - long division …), common, simplifying, rationalize … 𝑥 2 +3𝑥−4 1+3(1)−4 0 Eg1: 𝑙𝑖𝑚 = = I.F The roots {1, – 4} 𝑥→1 𝑥−1 1−1 0 𝑥 2 +3𝑥−4 (𝒙−𝟏)(𝑥+4) (𝑥+4) 𝑙𝑖𝑚 = 𝑙𝑖𝑚 = 𝑙𝑖𝑚 =1+4=5 𝑥→1 𝑥−1 𝑥→1 (𝒙−𝟏) 𝑥→1 1 √𝑥−1−√3 √4−1 −√3 0 Eg2: 𝑙𝑖𝑚 = = I.F 𝑥→4 𝑥−4 4−4 0 √𝑥−1−√3 √𝑥−1−√3 √𝒙−𝟏+√𝟑 (𝑥−1−3) 𝑙𝑖𝑚 = 𝑙𝑖𝑚 × = 𝑙𝑖𝑚 (𝑥−4)( 𝑥→4 𝑥−4 𝑥→4 𝑥−4 √𝒙−𝟏+√𝟑 𝑥→4 √𝑥−1+√3) (𝒙−𝟒) 1 1 1 = 𝑙𝑖𝑚 (𝒙−𝟒)( = 𝑙𝑖𝑚 = = 𝑥→4 √𝑥−1+√3) 𝑥→4 (√𝑥−1+√3) √4−1+√3 2√3 (2𝑥 2 −3𝑥+1) 2−3+1 0 𝟏 Eg3: 𝑙𝑖𝑚 (𝑥−1) = = I. F The roots {1, } 𝑥→1 1−1 0 𝟐 (𝒙−𝟏)(𝟐𝑥−1) (2𝑥−1) 𝟏 𝑙𝑖𝑚 = 𝑙𝑖𝑚 = 2(1) – 1 = 1 Factors? x = → (2x – 1) 𝑥→1 (𝒙−𝟏) 𝑥→1 1 𝟐  11 Sc. Math Chapter 10 Page Adnan Lesson Limits of Functions at a point 4 Ashkar 𝑥 2 −3𝑥+2 1−3+2 0 Eg4: 𝑙𝑖𝑚 = = 𝐼. 𝐹 𝑥→1 (2𝑥−1)(√𝑥−1) (2−1)(√1−1) 0 𝒙𝟐 −𝟑𝒙+𝟐 √𝑥+1 (𝒙−𝟐)(𝒙−𝟏) √𝑥+1 𝑙𝑖𝑚 × = 𝑙𝑖𝑚 × 𝑥→1 (2𝑥−1)(√𝑥−1) √𝑥+1 𝑥→1 (2𝑥−1)(𝑥−1) 1 (𝒙−𝟐) √𝑥+1 1−2 √1+1 𝑙𝑖𝑚 × = × = –2 𝑥→1 (2𝑥−1) 1 2−1 1 3𝑥 3 −2𝑥 2 +5 3(−1)3 −2(−1)2 +5 0 Eg5: 𝑙𝑖𝑚 = = I.F 𝑥→ – 1 𝑥+1 −1+1 0 *: Divide P(x) by the factor (x + 1) 𝟑𝒙𝟐 − 5x + 5 x + 1 3𝒙𝟑 −2x2 + 5 ∓3𝒙𝟑 ∓ 3x2 −5x2 + 5 ± 5x2 ± 5x 5x + 5 ∓5x ∓ 5 *: Write the factored form of P(x) = (x + 1) (3x2−5x + 5) 0 (𝒙+𝟏)(3𝑥 2 −5𝑥+5) 𝑙𝑖𝑚 = 𝑙𝑖𝑚 (3𝑥 2 − 5𝑥 + 5) 𝑥→−1 (𝒙+𝟏) 𝑥→−1 = 3(–1)2 –5(–1) + 5 = 3 + 5 + 5 = 13 8 9 10 **Solve 129 129 129 3 - Vertical Asymptote (VA) If 𝒍𝒊𝒎𝒇(𝒙) = ± ∞ → the straight line (d) of equation x = a is a VA of graph f(x). 𝒙→𝒂 1 Eg1: Let f(x) =. Find 𝑙𝑖𝑚− 𝑓(𝑥) , 𝑙𝑖𝑚+ 𝑓(𝑥) , deduce the asymptote. 𝑥−3 𝑥→3 𝑥→3 3– 3+ x –∞ 3 +∞ x – 3 – + 1 1 𝑙𝑖𝑚 = = −∞ 𝑥→3− 𝑥−3 0− 1 1 𝑙𝑖𝑚+ = =+∞ 𝑥→3 𝑥−3 0+ Then x = 3 is a VA.  11 Sc. Math Chapter 10 Page Adnan Lesson Limits of Functions at a point 5 Ashkar 𝑥 2 +3𝑥−4 Eg2: Consider the function f defined on R – {– 1, 1} by f(x) = …(C) 𝑥 2 −1 Determine the limits of f at the boundaries of its domain & write a conclusion. (Find the HA, VA & …) D = R – {– 1, 1}  D = ] – ∞ , – 1 [  ] – 1 , 1 [ ] 1 ; + ∞ [ -1– -1+ 1– 1+ x –∞ –1 1 +∞ x2 − 1 + – + 𝑥 2 +3𝑥−4 𝑥2 𝑙𝑖𝑚 𝑓(𝑥) = 𝑙𝑖𝑚 = 𝑙𝑖𝑚 =1 𝑥→−∞ 𝑥→−∞ 𝑥 2 −1 𝑥→−∞ 𝑥 2 2 𝑥 +3𝑥−4 𝑥2 Then y = 1 is a HA at ± ∞ 𝑙𝑖𝑚 𝑓(𝑥) = 𝑙𝑖𝑚 = 𝑙𝑖𝑚 =1 𝑥→+∞ 𝑥→+∞ 𝑥 2 −1 𝑥→+∞ 𝑥 2 𝑥 2 +3𝑥−4 1+3(−1)−4 −6 𝑙𝑖𝑚− 𝑓(𝑥) = 𝑙𝑖𝑚− = = =–∞ 𝑥→−1 𝑥→−1 𝑥 2 −1 0+ 0+ Then x = – 1 is a VA 𝑥 2 +3𝑥−4 1+3(−1)−4 −6 𝑙𝑖𝑚+ 𝑓(𝑥) = 𝑙𝑖𝑚+ = = =+∞ 𝑥→−1 𝑥→−1 𝑥 2 −1 0− 0− 𝑥 2 +3𝑥−4 1+3−4 0 𝑙𝑖𝑚 𝑓(𝑥) = 𝑙𝑖𝑚 = = I.F 𝑥→+1 𝑥→+1 𝑥 2 −1 1−1 0 Then (1 ; 2.5) is a limiting point. (𝒙−𝟏)(𝑥+4) (𝑥+4) 1+4 𝑙𝑖𝑚 = 𝑙𝑖𝑚 = = 2.5 𝑥→+1 (𝒙−𝟏)(𝑥+1) 𝑥→+1 ((𝑥+1) 1+1 11 12 **Solve ++ a–, a+ (1+2) 129 130  11 Sc. Math Chapter 10 Page Adnan Lesson Limits of Functions at a point 6 Ashkar 4 - Continuity / Continuous functions a - Continuity at a point f is continuous at a if 𝑙𝑖𝑚−𝑓(𝑥) = 𝑙𝑖𝑚+𝑓(𝑥) = f(a). 𝑥→𝑎 𝑥→𝑎 Note: 𝒍𝒊𝒎 = 𝒍𝒊𝒎− & 𝒍𝒊𝒎 = 𝒍𝒊𝒎+ 𝒙→ 𝒂 𝒙→𝒂 𝒙→ 𝒂 𝒙→𝒂 𝒙< 𝒂 𝒙> 𝒂 Eg1: In the figure to the right: 𝑙𝑖𝑚+𝑓(𝑥) = 6 𝑥→2 𝑙𝑖𝑚 𝑓(𝑥) = 4 𝑥→2− f(2) = 5. Then f is not continuous at 2. b - Continuity on an interval * f is continuous on interval I if f is continuous on 𝒙, for all 𝒙 ∈ 𝑰. Refer to Eg1: 𝑓 is continuous on [3, 4]. 𝑓 is defined on [1, 4] and continuous on [1,2[∪ [2,4]. 2𝑥 − 3 𝑖𝑓 𝑥 ≤ 1 Eg2: Prove that f is continuous at x = 1 if f(x) = { −𝑥 𝑖𝑓 𝑥 > 1 𝑙𝑖𝑚 f(x) = 𝑙𝑖𝑚 f(x) = 𝑙𝑖𝑚− (2x – 3) = 2(1) – 3 = – 1 𝑥→ 1− 𝑥→ 1 𝑥→ 1 𝑥< 1 f(1) = 2(1) – 3 = – 1 𝑙𝑖𝑚 f(x) = 𝑙𝑖𝑚 f(x) = 𝑙𝑖𝑚+ (– x) = – 1. 𝑥→ 1+ 𝑥→ 1 𝑥→ 1 𝑥> 1 Then f is continuous at 1. **Eg3: Is f continuous at 1? If f(x) = x2 + 2x if x < 1 2 if x = 1 3x if x > 1 𝑙𝑖𝑚 f(x) = 𝑙𝑖𝑚− x2 + 2x = 12 + 2(1) = 3 𝑥→ 1− 𝑥→ 1 f(1) = 2 NOT 𝑙𝑖𝑚+f(x) = 3(1) = 3 𝑥→ 1 *Ans: 3 = 3 ≠ 2 → No  11 Sc. Math Chapter 10 Page Adnan Lesson Limits of Functions at a point 7 Ashkar **Eg4: Is f continuous at 0? If f(x) = x2 + 1 if x < 0 1 if x = 0 √𝑥 + 1 if x > 0 𝑙𝑖𝑚 f(x) = 0 + 1 = 1 𝑥→ 0− f(0) = 1 𝑙𝑖𝑚 f(x) = √0 + 1 = 1 𝑥→ 0+ *Ans: 1 = 1 = 1 → yes Note: If f is NOT defined at ‘a”, then it is NOT continuous at ‘a”. A polynomial function is The rational function is The irrational function is continuous on ℝ. continuous on its domain continuous on its domain of definition. of definition. 𝑓(𝑥) = 𝑥 2 1 𝑓(𝑥) = √𝑥 𝑓(𝑥) = 𝑥 𝑓 is defined on ℝ 𝑓 is defined on ℝ − {0} 𝑓 is defined on [0, +∞[ 𝑓 is continuous on ℝ 𝑓 is continuous on ℝ − {0} 𝑓 is continuous on [0, +∞[ 5 - Discontinuous: has breaks or jumps on its domain. f is discontinuous at 1 because we have jump/break at 1. 1  11 Sc. Math Chapter 10 Page Adnan Lesson Limits of Functions at a point 8 Ashkar Reading y 1: Df = R – { – 1, 1} A the n graph 2: 𝑙𝑖𝑚 f(x) = 1 s 𝑥→ −∞ w e 4 3: 𝑙𝑖𝑚 f(x) = 1 r 𝑥→ +∞ * asymptote: y = 1 is HA at ± ∞ 3 4: 𝑙𝑖𝑚 −f(x) = +∞ 𝑥→ −1 2 5: 𝑙𝑖𝑚 +f(x) = + ∞ 𝑥→ −1 * asymptote: x = – 1 VA 1 6: 𝑙𝑖𝑚−f(x) = –∞ 𝑥→ 1 -2 -1 0 1 2 3 4 x 7: 𝑙𝑖𝑚+f(x) = +∞ 𝑥→ 1 * asymptote: x = 1 VA 8: 𝑙𝑖𝑚−f(x) = 3 𝑥→ 2 y' 9: 𝑙𝑖𝑚+ f(x) = 2 𝑥→ 2 10: f(2) = 2.5 11: Is f continuous at 2? NO 3 ≠ 2 ≠ 2.5 OR….. ** Practice reading the graph. The adjacent figure represents the curve (C) 2016 9 y of a function f. 1124 8 q5 7 1: Determine the domain of definition of f. 6 2: Calculate the following limits: 5 𝑙𝑖𝑚 𝑓(𝑥), 𝑙𝑖𝑚 𝑓(𝑥), 4 𝑥→−∞ 𝑥→+∞ 𝑙𝑖𝑚 − 𝑓(𝑥), 𝑙𝑖𝑚 +𝑓(𝑥), 3 𝑥→(−1) 𝑥→(−1) 2 𝑙𝑖𝑚 𝑓(𝑥) , 𝑙𝑖𝑚+ 𝑓(𝑥). 𝑥→2− 𝑥→2 1 x 3: Write the equations of the asymptotes -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 to (C). -1 4: Determine two values of x for which f is not -2 continuous & justify. -3 -4 -5 1: Df = R – { – 1, 2} Answer 2: 𝑙𝑖𝑚 f(x) = 2 ; 𝑙𝑖𝑚 f(x) = 2 ; 𝑙𝑖𝑚 𝑓(𝑥) = + ∞ ; 𝑙𝑖𝑚 𝑓(𝑥) = − ∞ 𝑥→ −∞ 𝑥→ +∞ 𝑥→(−1)− 𝑥→(−1)+ 𝑙𝑖𝑚 𝑓(𝑥) = −∞ 𝑙𝑖𝑚 𝑓(𝑥) = +∞ 𝑥→2− 𝑥→2+ 3: y = 2 HA at ; x = − 1 VA ; x = 2 VA. 4: x = − 1 ; x = 2 because f is NOT defined at them. 11 Sc. Math Chapter 10 Page Adnan Lesson Limits of Functions at a point 9 Ashkar # WSH … I.H.S …. Exams (2016…2019) Date 1 Calculate the following limits & deduce the asymptotes when they exist: 2016 1124 √2𝑥+1−1 5𝑥+|𝑥| 3𝑥 2 −7𝑥+2 3:𝑙𝑖𝑚 4: 𝑙𝑖𝑚− 5: 𝑙𝑖𝑚 q1 𝑥→0 3𝑥 𝑥→0 3𝑥−|𝑥| 𝑥→2 𝑥 3 −8 2 5𝑥 2 +1 2016 Consider the function f defined by f(x) = 2 of curve (C). 1124 2𝑥 +𝑥−3 q2 1: Find the domain of definition of f. 2: Find the limits of f at the boundaries of its domain. 3: Deduce the equations of the asymptotes to the curve (C). 3 3𝑥−1 2016 𝑖𝑓 𝑥≤1 1124 𝑥−4 f is a function defined over ]−∞; 2[ ∪ ]2; +∞[ by 𝑓(𝑥) = {𝑥2 +1 q3 𝑖𝑓 𝑥>1 𝑥−2 Is f continuous at x = 1? 4 Study the continuity of f at x = 1, if the function f is defined by: 2016 𝑥−1 1205 f(x) = if x  [– 3, 1[ q1 √ 𝑥+3−2 4 if x = 1 𝑥 2 −2𝑥−8 if x]1, +∞ [ 𝑥−1 5 −4𝑥+20 𝑥 3 −1 2017 Find the limits & write your conclusion: 3: 𝑙𝑖𝑚 4: 𝑙𝑖𝑚 1120 𝑥→−5 𝑥 2 +10𝑥+25 𝑥→1 2𝑥 2 +𝑥−3 qa 6 𝒙𝟑 −𝟐𝟕 √𝒙+𝟒−𝟏 2017 Calculate the following limits: b: 𝑙𝑖𝑚 c: 𝑙𝑖𝑚 1127 𝑥→ 3 𝒙𝟐 −𝟗 𝑥→−3 𝒙+𝟑 q1 7 𝒙𝟐 −𝟐𝒙 2017 Given the function f such that f(x) = 𝟐 1127 𝟐𝒙 −𝟓𝒙+𝟑 q2 1: Determine the domain of definition of f. 2: Determine the limits of f at the boundaries of the domain. Deduce the asymptotes.  11 Sc. Math Chapter 10 Page Adnan Lesson Limits of Functions at a point 10 Ashkar Summary 1 - Definition 𝐿𝑖𝑚𝑓(𝑥) = k (where a R, kR & k: limit of function f at a). 𝑥→𝑎 * If k = ± ∞. Then f(x) has no limit. ** If 𝑙𝑖𝑚−𝑓(𝑥) = 𝑙𝑖𝑚+𝑓(𝑥) = k → 𝑙𝑖𝑚𝑓(𝑥) = k 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 2 - To avoid indeterminate forms of limits at a point: modify the expression by factoring (Identities – /using the calculator, - long division …), common, simplifying, rationalize … 3 - Vertical Asymptote (VA) If 𝑙𝑖𝑚𝑓(𝑥) = ± ∞ → (d) of equation x = a is a VA of graph f(x). 𝑥→𝑎 4 - Continuity / Continuous functions a: f is continuous at a if: 𝑙𝑖𝑚−𝑓(𝑥) = 𝑙𝑖𝑚+𝑓(𝑥) = f(a). 𝑥→𝑎 𝑥→𝑎 Note: 𝑙𝑖𝑚 = 𝑙𝑖𝑚− & 𝑙𝑖𝑚 = 𝐿𝑖𝑚+ 𝑥→ 𝑎 𝑥→𝑎 𝑥→ 𝑎 𝑥→𝑎 𝑥< 𝑎 𝑥> 𝑎 b: f is continuous on interval I if f is continuous on 𝑥, for all 𝑥 ∈ 𝐼. 5 - Discontinuous: has breaks or jumps on its domain. THANK YOU Adnan AL-Ashkar