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Limits at In…nity Lecture Notes page 1 Sample Problems 1. Compute each of the following limits. a) lim 3x4 b) 2x5 c) lim x!1 x!1 lim 3x4 d) x! 1 x!1 2x5 lim x! 1 2 6 x 3 e) lim f) g) lim 4x3 x!1 2 6 x 3 lim x! 1 lim 4x3 h) x! 1 2. Compute each of the following limits. a) lim 2x c) lim x!1 b) x!1 lim 2x d) x! 1 x 2 3 2 3 lim x! 1 2x+3 x!1 3x+1 2x+3 f) lim x+1 x! 1 3 22x+1 x!1 3x 1 22x+1 h) lim x! 1 3x 1 e) lim x g) lim 3. Compute each of the following limits. a) 1 x!1 x b) lim lim d) 1 1x x! lim x! 1 x!1 5 x!1 2x3 f) 7+ 2x3 + 1 e) lim c) lim 5 2x3 3x lim x! 1 8 x g) lim 5 12 + x x4 h) lim 5x3 x!1 5x3 i) x 2x + 4 x3 x!1 2 2x + 4 x2 5x3 lim x!1 2x + 4 x4 4. Compute each of the following limits. a) lim x! 1 b) lim x!1 2x5 2x5 8x4 + 7x3 8x4 + 7x3 10 c) 10 2x5 + 8x6 lim x! 1 d) lim x!1 2x5 + 8x6 5. Compute each of the following limits. a) x + x2 6 1 6x + 5x2 + 2x3 x2 + 9 x!1 5x + 2x2 3 lim x! b) lim c) x3 1 3x2 lim x! 9x + 1 2x 15 Practice Problems 1. Compute each of the following limits. a) lim x!1 b) lim x! 1 3 15 x 8 3 15 x 8 1 8 x x!1 3 c) lim d) lim x! 1 c copyright Hidegkuti, Powell, 2009 1 8 x 3 e) lim 4x9 x!1 f) lim 4x9 x! 1 g) lim x!1 h) lim x! 1 7x10 7x10 Last revised: November 27, 2013 Limits at In…nity Lecture Notes page 2 2. Compute each of the following limits. 23x x!1 5x a) lim b) 23x 1 5x lim x! 22x+3 1 5x 1 1 d) 1 lim x! f) 22x+3 1 4x 1 22x+3 3x x!1 7x 2 lim x! 1 g) lim e) lim c) lim 1 2x+3 3x x!1 7x 2 22x+3 x!1 4x 1 22x+3 x!1 5x 1 1 1 h) lim 3. Compute each of the following limits. 3 x!1 x5 a) lim b) 3 1 x5 x! x!1 lim x! 1 3+ 5 x3 3+ x! 1 5x j) 1 x!1 lim x! 1 4x8 + x3 x4 x!1 3 p) 5 3x5 + 2x x2 lim o) lim x x! 1 l) n) 3 5x k) lim 7 6x x!1 x lim 3x5 + 2x x2 m) lim 2 x+3 5x x!1 7 6x 5 x3 lim i) lim 2 5 + 4 x 3x 1 x! 1 x!1 f) 2 5 + 4 x 3x 1 lim e) lim h) 2 x+3 5x x!1 lim c) lim d) g) lim lim x! 1 5 x+7 4x8 + x3 x4 x+7 3x 2x 1 x! 1 3x 2x 4. Compute each of the following limits. a) lim x! 1 b) lim x!1 7x5 + x3 c) 7x5 + x3 lim x! 1 d) lim x!1 1 6 x 4 120x5 120x5 1 6 x 4 e) f) lim x! 1 lim x! 1 8x4 8x4 1 x+2 5 3x3 1 x+2 5 3x3 5. The graph of a polynomial function is shown on the picture below. What can we state about this polynomial based on its end-behavior? y x 6. Compute each of the following limits. a) b) 1 1x lim x! 5 1 2x3 lim x! c) lim x! 1 d) lim x!1 3 c copyright Hidegkuti, Powell, 2009 2 5 x3 2 11 + x x4 2x2 + 3x + 1 x!1 3x2 5x + 2 e) lim f) lim x! 1 3x3 + 2x + 1 5x 3 g) h) lim x! 1 3x2 5x2 1 3x + 2 20x 2x2 42 1 5x3 20x2 105x lim x! Last revised: November 27, 2013 Limits at In…nity Lecture Notes page 3 Sample Problems - Answers 1. a) 1 b) 1 2. a) 1 b) 0 c) 1 c) 0 3. a) 0 b) 0 4. a) 1 b) 5. a) 0 b) d) 1 d) 1 c) 0 1 d) c) 1 e) 0 7 c) 1 1 2 e) f) f) 1 e) 1 g) 1 g) 1 1 f) 3 h) 1 h) 0 g) 1 h) 5 i) 0 d) 1 1 Practice Problems - Answers 1. a) 1 2. a) 1 b) 0 3. a) 0 m) b) 1 b) 0 1 b) c) 1 1 lim f (x) = x! 1 coe¢ cient. 6. a) 0 c) 0 n) 1 4. a) 1 5. Since c) 1 b) 0 d) 1 d) 1 d) 1 o) e) 32 e) 3 1 c) e) 1 p) 1 f) 1 f) 32 f) 3 g) 1 1 e) 1 g) g) 0 h) 1 h) 1 h) 1 1 i) 5 j) 5 k) 3 2 l) 3 2 1 d) f) 1 1 and lim f (x) = 1, the polynomial is of odd degree and has a positive leading x!1 c) 2 c copyright Hidegkuti, Powell, 2009 d) 3 e) 2 3 f) 1 g) 3 5 h) 0 Last revised: November 27, 2013 Limits at In…nity Lecture Notes page 4 Sample Problems - Solutions 1. Compute each of the following limits. a) lim 3x4 x!1 Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large positive number. Then 3x4 is very large, and also positive because it is the product of …ve positive numbers. 3x4 = 3 x positive x positive positive x positive x positive So the answer is 1. We state the answer: lim 3x4 = 1. x!1 b) lim 3x4 x! 1 Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a very large negative number. Then 3x4 is very large, and also positive because it is the product of one positive and four negative numbers. 3x4 = 3 positive x!1 x negative x negative x negative lim 3x4 = 1 So the answer is 1. We state the answer: c) lim x negative x! 1 2x5 Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large positive number. Then 2x5 is very large, and also negative because it is the product of one negative and …ve positive numbers. 2x5 = 2 x x x x x positive negative So the answer is d) lim x! 1 positive positive 2x5 = 1. We state the answer: lim x!1 positive positive 1 2x5 Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a very large negative number. Then 2x5 is very large, and also positive because it is the product of six negative numbers. 2x5 = 2 x x x x x negative negative So the answer is 1. We state the answer: e) lim x!1 negative negative negative negative 2x5 = 1 lim x! 1 2 6 x 3 Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large 2 6 positive number. Then x is very large, and also negative because it is the product of one negative and 3 six positive numbers. 2 6 x = 3 c copyright Hidegkuti, Powell, 2009 2 3 x positive x positive x positive x positive x positive x positive negative Last revised: November 27, 2013 Limits at In…nity Lecture Notes So the answer is 1. We state the answer: 2 6 x 3 lim x!1 = page 5 1 2 6 x x! 1 3 Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a 2 6 x is very large, and also negative because it is the product of seven very large negative number. Then 3 negative numbers. 2 6 2 x = x x x x x x 3 3 negative negative negative negative negative negative f) lim negative So the answer is 1. We state the answer: 2 6 x = 3 lim x! 1 1 g) lim 4x3 x!1 Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large positive number. Then 4x3 is very large, and also positive because it is the product of four positive numbers. 4x3 = 4 positive x x positive x positive positive So the answer is 1. We state the answer: lim 4x3 = 1 x!1 h) lim 4x3 x! 1 Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a very large negative number. Then 4x3 is very large, and also negative because it is the product of one positive and three negative numbers. 4x3 = 4 x x x positive So the answer is 1. We state the answer: negative lim x! 1 4x3 negative = negative 1 2. Compute each of the following limits. Let a > 0: Then the limit of the exponential function f (x) = ax is as follows. lim ax = 1 and Case 1. If a > 1, then Case 2. If 0 < a < 1, then lim ax = 0 x!1 x! 1 lim ax = 0 and x!1 y -5 -4 -3 -2 -1 0 1 lim ax = 1 x! 1 y 2 3 4 5 -5 -4 -3 -2 -1 0 1 x a>1 c copyright Hidegkuti, Powell, 2009 2 3 4 5 x 0<a<1 Last revised: November 27, 2013 Limits at In…nity Lecture Notes a) lim 2x and x!1 page 6 lim 2x b) x! 1 Solution: Since 2 > 1; these limits are 1 and 0; i.e. lim 2x = 1 and lim 2x = 0. x!1 c) lim x!1 2 3 x and Solution: Since d) lim x! 1 x! 1 x 2 3 2 < 1; these limits are 0 and 1, i.e. lim x!1 3 2 3 x = 0 and lim x! 1 2x+3 e) lim x+1 x!1 3 Solution: We start by re-writing the exponential expressions. there is only one exponential expression involving x: 2 3 x f) 2x+3 1 3x+1 2x+3 2x+3 8 = lim = lim 1 3x+1 x! 1 3x+1 x! 1 3 2 3 2x+3 8 = lim x+1 x!1 3 x!1 3 lim 8 lim 3 x!1 = = 1. 2 3 x x 2 3 =0 since 2 <1 3 lim x! Solution: lim x! x = 8 lim 3 x! 1 22x+1 g) lim x 1 x!1 3 Solution: We start by re-writing the exponential expressions. there is only one exponential expression involving x: 2 3 x =1 The goal is to bring it into a form where x 22 2 22x+1 4x 6 22x 21 = = = =6 x 1 3 3x 1 3x x 3 3 31 Thus h) x The goal is to bring it into a form where 2x 23 2x 8 8 2x+3 = = = 3x+1 3x 31 3x 3 3 Thus 2 3 22x+1 = lim 6 x!1 3x 1 x!1 lim 22x+1 1 3x 1 4 3 x = 6 lim x!1 4 3 4 3 x x =1 since 4 >1 3 lim x! Solution: 22x+1 = lim 6 1 3x 1 x! 1 lim x! 4 3 x = 6 lim x! 1 4 3 x =0 3. Compute each of the following limits. 1 a) lim x!1 x Solution: This is a very important limit. Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large positive number. The reciprocal of a very large positive number is a very small positive number. This limit is 0. 1 x Solution: Since the limit we are asked for is as x approaches negative in…nity, we should think of x as a very large negative number. The reciprocal of a very large negative number is a very small negative number. This limit is 0. b) lim x! 1 c copyright Hidegkuti, Powell, 2009 Last revised: November 27, 2013 Limits at In…nity Lecture Notes page 7 5 2x3 Solution: Since the limit we are asked for is as x approaches in…nity, we should think of x as a very large positive number. We divide 5 by a very large positive number. This limit is 0. c) lim x!1 d) lim x! 1 5 2x3 7+ 8 x Solution: This limit is 7 since the other two terms aproach zero as x approaches negative in…nity. Using mathematical notation, 5 1 2x3 7+ lim x! x!1 7+0= 7 5 12 + x x4 2x3 + 1 e) lim 8 5 8 = lim + lim 7 + lim =0 3 x! 1 x! 1 x x x! 1 2x Solution: This limit is 1 since the …rst term approaches negative in…nity, the second term approaches 1 and the other two terms aproach zero as x approaches in…nity. Using mathematical notation, lim x!1 f) lim 5 12 + x x4 2x3 + 1 3x 5 x 2x3 + lim 1 + lim = lim x!1 x!1 x!1 12 x4 + lim x!1 = 1+1+0+0= 1 2 x Solution: This problem is similar to the previous problems after a bit of algebra. We simply divide by x and then the limit becomes familiar. x! 1 lim 3x g) lim 5x3 2 x x! 1 = lim x! 1 3x x 2 x = lim x! 1 3 2 x =3 2x + 4 x2 x!1 Solution: 5x3 lim x2 x!1 h) lim 2x + 4 5x3 5x3 2x 4 2 4 + 2 + 2 = lim 5x + 2 2 x!1 x!1 x x x x x 2 4 = lim ( 5x) + lim + lim = 1+0+0= x!1 x!1 x!1 x x2 = lim 1 2x + 4 x3 x!1 Solution: 5x3 lim x3 x!1 i) lim x!1 5x3 2x + 4 = lim 2x 4 5x3 + 3 + 3 3 x x x = lim 5 = lim 2x 4 5x3 + 4 + 4 4 x x x = lim 5 x x!1 x!1 4 2 + 3 2 x x = 5 2x + 4 x4 Solution: lim x!1 5x3 2x + 4 x4 c copyright Hidegkuti, Powell, 2009 x!1 x!1 2 4 + 4 3 x x =0 Last revised: November 27, 2013 Limits at In…nity Lecture Notes page 8 4. Compute each of the following limits. a) lim x! 1 2x5 8x4 + 7x3 10 Solution: The …rst term, 2x5 approaches in…nity and the secomd term, 8x4 approaches negative in…nity. This does not give us enough information about the entire polynomial. A limit like this is called an indeterminate. We will bring this expression to a form that is not an indeterminate. In this case, factoring out the …rst term does the trick. In case of a polynomial, the limits at in…nity and negative in…nity are completely determined by its leading term. Recall that the leading term is the highest degree term. 2x5 lim x! 1 8x4 + 7x3 2x5 10 = lim x! 1 Here is the computation showing why this is true. We …rst factor out the entire leading term. lim x! 1 2x5 8x4 + 7x3 10 = = = lim 2x5 lim 2x5 lim 2x5 x! 1 x! 1 x! 1 4 x 1+ lim x! 1 7 5 + 5 2 x x 4 7 5 1+ + x x2 x5 2x5 1 = lim x! 1 We can now easily determine that this limit is 1. (See problem number 1.) b) lim x!1 2x5 8x4 + 7x3 10 Solution: In case of a polynomial, the limits at in…nity and negative in…nity are completely determined by its leading term. Recall that the leading term is the highest degree term. 2x5 lim x!1 8x4 + 7x3 2x5 10 = lim x!1 Here is the computation showing why this is true. We …rst factor out the entire leading term. lim x!1 2x5 8x4 + 7x3 10 = = = We can now easily determine that this limit is c) lim x! 1 lim 2x5 lim 2x5 lim 2x5 x!1 x!1 x!1 4 x 1+ lim x!1 7 5 + 2x2 x5 7 5 4 1+ + 5 2 x 2x x 2x5 1 = lim x!1 1. (See problem number 1.) 2x5 + 8x6 Solution: In case of a polynomial, the limits at in…nity and negative in…nity are completely determined by its leading term. lim 2x5 + 8x6 = lim 2x5 + 8x6 = x! 1 x! 1 = lim 8x6 = 1 because x! 1 lim 8x6 lim 6 x! 1 x! 1 8x 2x5 = lim x! 1 6 8x6 1 1 4x = lim x! 1 8x6 lim x! 1 1 1 4x 1 = lim 8x x! 1 We can now easily determine that this limit is 1. (See problem number 1.) c copyright Hidegkuti, Powell, 2009 Last revised: November 27, 2013 Limits at In…nity Lecture Notes page 9 2x5 + 8x6 d) lim x!1 Solution: In case of a polynomial, the limits at in…nity and negative in…nity are completely determined by its leading term. lim 2x5 + 8x6 = lim 2x5 + 8x6 = x!1 x!1 lim 8x6 = 1 because x!1 lim 8x6 2x5 = lim 8x6 lim 8x6 1=1 x!1 = x!1 1 4x 1 x!1 = lim 8x6 x!1 lim x!1 1 1 4x We can now easily determine that this limit is 1. (See problem number 1.) 5. Compute each of the following limits. x + x2 6 =0 x! 1 6x + 5x2 + 2x3 Solution: The numerator approaches in…nity and the denominator approaches negative in…nity. This does not give us enough information about the quotient. A limit like this is called an indeterminate. We will bring this expression to a form that is not an indeterminate. Let us rearrange the polynomials in the rational function given. Then we will factor out the leading term in the numerator and denominator. a) lim x2 + x 6 lim = lim x! 1 2x3 + 5x2 + 6x x! 1 1 6 x x2 5 3 1+ + 2 2x x x2 1 + 2x3 We now express the limit of the product as the product of two limits 1 6 x x2 5 3 1+ + 2 2x x x2 1 + lim x! 1 2x3 = lim x! 1 1 6 x x2 5 3 1+ + 2 2x x 1+ x2 2x3 lim x! 1 The …rst expression can be simpli…ed and thus has a limit we can easily determine its limit. The second expression, although looks unfriendly, is always going to approach 1. x2 lim x! 1 2x3 6 1 x x2 5 3 1+ + 2 2x x 1+ lim x! 1 1 1=0 1=0 1 2x = lim x! The entire computation should look like this: x2 + x 6 1 2x3 + 5x2 + 6x lim x! = = c copyright Hidegkuti, Powell, 2009 1 6 x x2 5 3 1+ + 2 2x x x2 1 + lim x! 1 2x3 1 1=0 1=0 1 2x x2 1 2x3 = lim x! 1 6 1 1+ B x x2 C lim @ 5 3 A x! 1 1+ + 2 2x x 0 lim x! Last revised: November 27, 2013 Limits at In…nity Lecture Notes page 10 x2 + 9 1 = 2 x!1 5x + 2x 3 2 Solution: Both numerator and denominator approach in…nity. This does not give us enough information about the quotient. A limit like this is called an indeterminate. We will bring this expression to a form that is not an indeterminate. Let us rearrange the polynomials in the rational function given. Then we will factor out the leading term in the numerator and denominator. b) lim x2 1 + x2 + 9 lim x!1 2x2 + 5x 3 x3 x! 1 3x2 Solution: c) lim lim x! 1 9x + 1 = 2x 15 x3 3x2 9x + 1 2x 15 9 x2 1+ x2 = lim x!1 2x2 lim 5 x!1 3 1+ 2 2x 2x 9 1+ 2 1 1 1 x = lim lim = 1= 5 3 x!1 2 x!1 2 2 1+ 2x 2x2 = lim x!1 2x2 5 1+ 2x 9 x2 3 2x2 1 x3 1 = = lim x! 1 x3 1 3x2 lim x! 3x2 1 9 + x2 2 3x 1= 1 x3 5 x2 x3 = lim x! 1 3x2 x 1 3 lim x! 1= 1 lim x! 1 1 1= 1 9 + x2 2 3x 1 x3 5 x2 1 For more documents like this, visit our page at http://www.teaching.martahidegkuti.com and click on Lecture Notes. E-mail questions or comments to [email protected]. c copyright Hidegkuti, Powell, 2009 Last revised: November 27, 2013