Questions and Answers #5 Limits and Continuity

Questions and Answers Sheet 5 Limits and Continuity Question #1 Determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are ∞ or −∞ , and describe the corresponding vertical asymptote. lim sin 𝑥 𝑥→0± |𝑥| Answer: The given limit is lim± 𝑥→0 sin 𝑥 |𝑥| Obtain the right hand limit as shown below: lim sin 𝑥 𝑥→0+ |𝑥| = lim+ sin 𝑥 𝑥 𝑥→0 =1 Obtain the left hand limit as shown below: lim sin 𝑥 𝑥→0− |𝑥| = lim− 𝑥→0 sin 𝑥 −𝑥 =-1 Question #2 Determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are ∞ or −∞, and describe the corresponding vertical asymptote. lim 𝑥−sin|𝑥| 𝑥3 𝑥→0± Answer: To evaluate: lim+ 𝑥−sin|𝑥| 𝑥→0 𝑥3 We know that, − x if | x |=   x if x0 x0 Therefore, lim 𝑥→0+ It is 𝑥−sin|𝑥| 𝑥3 0 0 = lim 𝑥−sin(𝑥) 𝑥3 𝑥→0 form. Using L-hospital rule. lim 𝑥−sin 𝑥 𝑥3 𝑥→0 = lim 𝑥→0 𝑑 (𝑥−sin 𝑥) = lim 𝑑𝑥 𝑑 𝑥→0 𝑑𝑥 (𝑥 3 ) 1−cos 𝑥 3𝑥 2 𝑑 (1−cos 𝑥) 𝑑𝑥 𝑑 (3𝑥 2 ) 𝑥→0 𝑑𝑥 = lim = lim 𝑥→0 0+sin 𝑥 6𝑥 𝑑 (sin 𝑥) 𝑑𝑥 𝑑 (6𝑥) 𝑥→0 𝑑𝑥 = lim = lim = = cos 𝑥 𝑥→0 6 cos 0 1 6 6 Hence, required answer is 1 6 Question #3 Determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are ∞ or -∞, and describe the corresponding vertical asymptote. 𝜋 lim cos ( (𝑥 − [𝑥])) 2 𝑥→2± Answer: The floor function gives the result as the largest integer that is less than the number on which this function is applied. It is represented by the expression [x]. The left-side limit of a function at a point is the value of the function obtained at a point very close to the left side of the point. The right-side limit of a function at a point is the value of the function obtained at a point very close to the right side of the point. The left side limit of the given expression is calculated as follows: 𝜋 𝜋 lim cos ( (𝑥 − [𝑥])) = lim cos ( ((2 − ℎ) − ([2 − ℎ]))) 𝑥→2− 2 ℎ→0 2 𝜋 = lim cos ( (2 − ℎ − 1)) 2 ℎ→0 𝜋 = lim cos ( (1 − ℎ)) 2 ℎ→0 𝜋 = cos ( ) 2 =0 The right side limit of the given expression is calculated as follows: 𝜋 𝜋 lim+ cos ( (𝑥 − [𝑥])) = lim cos ( ((2 + ℎ) − ([2 + ℎ]))) ℎ→0 𝑥→2 2 2 𝜋 = lim cos ( (2 + ℎ − 2)) 2 𝜋 ℎ→0 = lim cos ( ℎ) 2 ℎ→0 = cos(0) =1 Question #4 *Identify the location of any discontinuities of f (x) attached *For each discontinuity, determine the type of discontinuity by evaluating the left- and right-side limits. 𝑓(𝑥) = 𝑥−1 𝑥 2 +𝑥 Answer: To determine the nature of discontinuities of the given function. The function is not defined at the points x=0 and -1; (at all other points, it is continuous). The points of discontinuities are x=0 and x=-1 𝑥−1 𝑓(𝑥) = 2 = 𝑥−1 𝑥 +𝑥 𝑥(𝑥+1) (𝑥 ≠ 0, −1) Analyzing the left and right hand limits at x=0 Left hand limit 1 𝑥−1 𝑥 𝑥→0 𝑥+1 = lim− 𝑓(𝑥) = lim− ( ) × lim 𝑥→0 𝑥→0 = ∞ Right - hand limit 1 𝑥−1 lim = lim− ( ) × lim = −∞. 𝑥→0− 𝑥→0 𝑥→0 𝑥 + 1 𝑥 𝑥−1 (∴ lim = −1) 𝑥→0 𝑥 + 1 Nether limits exist, so f(x) has discontinuity of the second kind at x=0 Analyzing the discontinuity at x=-1 Left - hand limit = lim− 𝑓(𝑥) = lim− ( 𝑥→−1 𝑥→−1 1 𝑥+1 ) × lim 𝑥−1 𝑥→−1 𝑥 =− ∞ Right - hand limit 1 𝑥−1 = lim+ ( ) × lim =∞ 𝑥→−1 𝑥 𝑥→−1 𝑥+1 𝑥−1 (∴ lim = 2) lim 𝑥→−1+ 𝑥→−1 𝑥 Nether limits exist, so f(x) has discontinuity of the second kind at x=-1 ANSWER: Points of discontinuity are x=0 ,x=-1, both are discontinuities of the second kind (essential discontinuities) Question #5 Derivatives from limits. The following limits represent ƒ'(a) for some function f and some real number a. a. Find a possible function f and number a. b. Evaluate the limit by computing f'(a) lim 𝑥→0 𝑒 𝑥 −1 𝑥 Answer: Given, lim 𝑥→0 𝑒 𝑥 −1 𝑥 a). lim 𝑒 𝑥 −1 𝑥→0 𝑥 = lim 𝑒 𝑥 −𝑒 0 𝑥→0 𝑥−0 Now comparing with lim 𝑥→𝑎 𝑥 𝑓(𝑥)−𝑓(𝑎) 𝑥−𝑎 , we get 𝑓(𝑥) = 𝑒 and a=0 b). The derivative of a function f at x=a is, 𝑓 ′ (𝑎) = lim 𝑓(𝑥)−𝑓(𝑎) 𝑥−𝑎 𝑥→𝑎 Now differentiating 𝑓(𝑥) = 𝑒 𝑥 with respect to x, we get 𝑓 ′ (𝑥) = 𝑒 𝑥 Now at x=0 𝑓 ′ (0) = 𝑒 0 =1 Therefore, lim 𝑥→0 𝑒 𝑥 −1 𝑥 = lim 𝑒 𝑥 −𝑒 0 𝑥→0 𝑥−0 =f'(0) [By the definition of derivative] =1 Question #6 Find the following: lim ( 1 𝑥→−2 𝑥 2 + 1 𝑥 3 +4 ) Explain your answers using the continuity property of the addition, subtraction, multiplication, and division. Answer: Given, lim ( 1 𝑥→−2 𝑥 2 + 1 𝑥 3 +4 ) By addition property of limits we know lim [𝑓(𝑥) + 𝑔(𝑥)] 𝑥→𝑎 = lim 𝑓(𝑥) + lim 𝑔(𝑥) 𝑥→𝑎 𝑥→𝑎 Therefore lim ( 1 𝑥→−2 𝑥 2 + 1 𝑥 3 +4 ) = lim 1 𝑥→−2 𝑥 2 + lim 1 𝑥→−2 𝑥 3 +4 1 = lim + lim 𝑥→−2 𝑥 2 1 1 = + (−2)3 4 1 1 4 1 −8+4 1 4 4 = + 1 𝑥→−2 𝑥 3 +4 +4 = − =0 Question #7 Using: lim 𝑓(𝑥) = 3 and lim 𝑥→4 𝑔(𝑥) = 7 𝑥→4 Answer: Given, lim 𝑓(𝑥) = 3 and lim 𝑔(𝑥) = 7 𝑥→4 𝑥→4 Now, lim 𝑓(𝑥) = 3 𝑥→4 and lim 𝑔(𝑥) = 7 𝑥→4 ∴ lim 𝑓(𝑥) ⋅ 𝑔(𝑥) = lim 𝑓(𝑥) lim 𝑔(𝑥) 𝑥→4 𝑥→4 𝑥→4 = (3) ⋅ (7) =21 ∴ lim = 𝑓(𝑥) 𝑥→4 𝑔(𝑥) 3 lim 𝑓(𝑥) = 𝑥→4 lim 𝑔(𝑥) 𝑥→4 7 Answer: lim 𝑓(𝑥) ⋅ 𝑔(𝑥) = 21 𝑥→4 lim 𝑓(𝑥) = 𝑥→4 𝑔(𝑥) 3 7 Question #8 Find the limit 𝑙𝑖𝑚𝑥→7 𝑥 2 +2𝑥 𝑥 2 −41 and explain your answer using the continuity property of the addition, subtraction, multiplication, and division. Answer: Solving the limit we get, 𝑙𝑖𝑚𝑥→7 𝑥 2 +2𝑥 𝑥 2 −41 Substituting the variable x with 7, we get = = (7)2 +2(7) (72 )−41 49+14 63 49−41 = 8 Question #9 Calculating limits. Determine the following limits. lim (|𝑥 − 1| + 𝑥) and lim (|𝑥 − 1| + 𝑥) 𝑥→−∞ 𝑥→∞ Answer: Given lim (|𝑥 − 1| + 𝑥) and lim (|𝑥 − 1| + 𝑥) 𝑥→−∞ Now the limit lim (|𝑥 − 1| + 𝑥) 𝑥→−∞ 𝑥→∞ Here |x-1|=-(x-1) when 𝑥 → −∞ Now lim (|𝑥 − 1| + 𝑥) = lim (−(𝑥 − 1) + 𝑥) 𝑥→−∞ 𝑥→−∞ = lim (−𝑥 + 1 + 𝑥) 𝑥→−∞ = lim (1) 𝑥→−∞ =1 And the limit lim (|𝑥 − 1| + 𝑥) 𝑥→∞ Here |x-1|=(x-1) when 𝑥 → ∞ Now lim (|𝑥 − 1| + 𝑥) = lim ((𝑥 − 1) + 𝑥) 𝑥→∞ 𝑥→∞ = lim (𝑥 − 1 + 𝑥) 𝑥→∞ = lim (2𝑥 − 1) 𝑥→∞ = (2(∞) − 1) =∞ Final answer: lim (|𝑥 − 1| + 𝑥) = 1 and lim (|𝑥 − 1| + 𝑥) = ∞ 𝑥→−∞ 𝑥→−∞ Question #10 Evaluate the following limits. Some require Squeeze Theorem some do not. If you are using Squeeze Theorem, make sure it is clear how you bounded your function and why Squeeze Theorem applies. lim 3𝑥 4 𝑒 1 cos( 2 ) 𝑥 𝑥→0 Answer: In the question it is asked to calculate the limit of following function given below: lim 3𝑥 4 𝑒 1 cos( 2 ) 𝑥 𝑥→0 As cos 𝜃 has values between -1 to 1. we can write it as 1 −1 ≤ cos ( 2) ≤ 1 𝑥 multiplying it by e 1 𝑒^{−1} ≤ 𝑒^{cos ( 2) ≤ 𝑒 1 𝑥 now multiplying it by → 3𝑥 4 cos( 1 ) 𝑥2 ≤ 3𝑥 4 𝑒 1 3𝑥 3 𝑒 −1 ≤ 3𝑥 4 𝑒 as 𝑥 → 0; 3𝑥 4 𝑒 → 0and 3𝑥 4 𝑒 −1 → 0 So by squeeze theorem or sandwich theorem 𝑈𝑥→0 3𝑥 4 𝑒 1 cos( 2 ) 𝑥 =0 Question #11 Find the limits. if the limit does not exist, state it. In each part show your calculations and circle the answer. 𝑥 2 −4𝑥+4 𝑥 2 +𝑥−6 Answer: Given: 𝑓(𝑥) = 𝑥 2 −4𝑥+4 𝑥 2 +𝑥−6 Left hand limit lim 𝑥→−3− 𝑓(𝑥) = lim − 𝑥→−3 𝑥 2 −4𝑥+4 𝑥 2 +𝑥−6 lim 𝑓(𝑥) = ∞ 𝑥→−3− The function grows without bound. Right hand limit lim + 𝑓(𝑥) = lim + 𝑥→−3 𝑥 2 −4𝑥+4 𝑥→−3 lim 𝑥 2 +𝑥−6 𝑓(𝑥) = − ∞ 𝑥→−3+ The function decreases without bound. lim − 𝑓(𝑥) ≠ lim + 𝑓(𝑥) 𝑥→−3 𝑥→−3 Since the corresponding two-sided limits are not equal, lim 𝑥 2 −4𝑥+4 𝑥→−3 𝑥 2 +𝑥−6 does not exist. Question #12 Find the limits: 8x2 − 3 2 x2 + x lim x → Answer: Given: Consider rational function inside the radical. The highest power of x in the denominator is 2. So divide numerator and denominator by 𝑥 2 , 8x2 3 − 2 2 8x − 3 x x lim = lim 2 2 x → x → 2x x 2x + x + 2 2 x x 2 = lim 𝑥→∞√ 3 8− 2 𝑥 1 2+𝑥 Use the limit laws, 3 lim 𝑥→∞√ 8𝑥2 −3 2𝑥2 +𝑥 = √ lim (8− 2 ) 𝑥 1 𝑥→∞ (2+𝑥) 3 lim (8− 2 ) 𝑥 = √𝑥→∞ 1 𝑥 lim (2+ ) 𝑥→∞ =√ 8−0 2+0 = √4 =2 Therefore, lim 𝑥→∞√ 8𝑥2−3 2𝑥2+𝑥 =2 Question #13 Define the function: 1−arctan 𝑥 𝑓(𝑥) = sin(𝑥)−cos(𝑥) and compute lim 𝑓(𝑥). Once again, we have not given you directions on computing limits so use one of 𝑥→√3 the methods outlined above to access the help files and determine how to do this Answer: Solve, as follows: 𝑓(𝑥) = 1−tan−1 (𝑥) sin(𝑥) − cos(𝑥) 1 − tan −1 ( x) 3 sin( x0 lim f ( x) = lim x→ 3 x→ 1−tan−1 (√3) = lim 𝑥→√3 sin(√3)−cos(√3) Solve, as follows: lim 𝑓(𝑥) = 𝑥→∞ 𝜋 3 1− 0.987026−(−0.1605565) =-0.041127 Question #14 Calculating limits The following limits are the derivatives of a composite function g at a point a. a. Find a possible function g and number a. b. Use the Chain Rule to find each limit. Verify your answer by using a calculator lim √4+sin 𝑥−2 𝑥 𝑥→0 Answer: lim √4+sin 𝑥−2 𝑥 𝑥→0 a. It can be written as, lim √4+sin 𝑥−2 𝑥 𝑥→0 = lim √4+sin 𝑥−√4−sin 0 𝑥−0 𝑔(𝑥)−𝑔(𝑎) 𝑥→0 Now comparing with lim 𝑥→𝑎 𝑥−𝑎 , we get 𝑔(𝑥) = √4 + sin 𝑥 and a=0 b. Differentiating 𝑔(𝑥) = √4 + sin 𝑥 with respect to x, we get 𝑔′ (𝑥) = = = 1 2√4+sin 𝑥 1 2√4 sin 𝑥 cos 𝑥 ⋅ 𝑑(4+sin 𝑥) 𝑑𝑥 ⋅ (0 + cos 𝑥) 2√4+sin 𝑥 Therefore, lim √4+sin 𝑥−2 𝑥 𝑥→0 = lim √4+sin 𝑥−√4+sin 0 𝑥→0 𝑥−0 =g'(0) = = cos 0 2 4 + sin 0 1 4 By using calculator, we get lim 𝑥→0 √4+sin 𝑥−2 𝑥 = 1 4 Question #15 Use continuity and the intermediate value theorem to solve problems. If 𝑓(𝑥) = 𝑥 3 − 6𝑥 + 8, show that there are values c for which f(c) equals a) 𝜋 , b) −√3 , and c) 6,000,000. Answer: Given 𝑓(𝑥) = 𝑥 3 − 6𝑥 + 8 Note that intermediate value theorem states that if f is continuous on a domain containing an interval [a,b] and let be s be number with f(a)