Todays Agenda Reminder: Test 0 (on precalculus) is this Friday, - PowerPoint PPT Presentation

Today’s Agenda • Reminder: Test 0 (on precalculus) is this Friday, 8/28. If you do well, you will get extra credit toward your final course grade. For practice exams, you can go to https: //math.asu.edu/first-year-math/mat-170-precalculus . • Upcoming Homework • Section 1.3: The limit of a function • Section 1.4: Calculating limits Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 1 / 12

Upcoming Homework • WeBWorK #2: Section 1.3 due 8/28/2015 • WeBWorK #3: Section 1.4 due 8/31/2015 (this one is really long!) • Written HW A: Section 1.3, #6 (please use graph paper) and #18 (your table of values should include at least 10 entries - use powers of 10); Section 1.4, #10, 24, 26, 38, 44 (for #24, 26, and 38, describe in a sentence or two the method you used to find the limit). Due 9/2/2015. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 2 / 12

Section 1.3 Last class period, we started the following problems. Let’s pick up where we left off: take a few moments to discuss the following problems with the person next to you. You can use a table on your graphing calculator, or an actual graph of the function in question, to help you determine the limits. 1 x − 1 lim x 2 − 1 x → 1 2 √ t 2 + 9 − 3 lim t 2 t → 0 3 � π � x → 0 sin lim x Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 3 / 12

Section 1.3 Definition 1.3.2 We write x → a − f ( x ) = L lim and say that the left-hand limit as x approaches a is equal to L if we can make the values of f ( x ) arbitrarily close to L by taking x to be sufficiently close to a and x < a . Similarly, if we require that x > a , we get the right-hand limit of f ( x ) as x approaches a , and we denote this by x → a + f ( x ) = M . lim It is very important to note that even if both the right- and left-hand limits exist, they do not have to equal each other! Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 4 / 12

Section 1.3 Theorem 1.3.3 Assuming we have a function f : D → R such that f is defined on an open interval containing a (except for possibly at a itself), the following is true: lim x → a f ( x ) = L if and only if x → a − f ( x ) = L lim and x → a + f ( x ) = L . lim Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 5 / 12

Section 1.3 Let x 2 ,  x < 1  f ( x ) = 2 x − 1 , 1 < x ≤ 5 10 , x > 5  With the person next to you, calculate the following: 1 the domain of f 2 lim x → 1 − f ( x ) 3 lim x → 1 + f ( x ) 4 lim x → 1 f ( x ) 5 lim x → 5 + f ( x ) 6 lim x → 5 − f ( x ) 7 lim x → 5 f ( x ) Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 6 / 12

Section 1.4 It turns out that limits are very well-behaved in terms of sums, differences, multiplication by constants, products, and quotients. Limit Laws Suppose that c is a constant, and that the limits lim x → a f ( x ) and lim x → a g ( x ) exist. Then 1 lim x → a [ f ( x ) + g ( x )] = lim x → a f ( x ) + lim x → a g ( x ) 2 lim x → a [ f ( x ) − g ( x )] = lim x → a f ( x ) − lim x → a g ( x ) 3 lim x → a [ c · f ( x )] = c · lim x → a f ( x ) 4 lim x → a [ f ( x ) · g ( x )] = lim x → a f ( x ) · lim x → a g ( x ) f ( x ) g ( x ) = lim x → a f ( x ) 5 lim x → a lim x → a g ( x ), provided that lim x → a g ( x ) � = 0 Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 7 / 12

Section 1.4 Using these limit laws, we can derive even more limit laws: 6. lim x → a [ f ( x )] n = [lim x → a f ( x )] n , when n is a positive integer 7. lim x → a c = c (the limit of a constant function f ( x ) = c is simply equal to the constant) 8. lim x → a x = a 9. lim x → a x n = a n √ x = √ a 10. lim x → a n n � � 11. lim x → a n f ( x ) = n lim x → a f ( x ) (provided that both sides are defined) In summary, limit laws are easy to remember because they do exactly what you think they should do. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 8 / 12

Section 1.4 Theorem 1.4.1 If f ( x ) ≤ g ( x ) for all x in some open interval containing a (except for possibly at a itself), and the limits of f and g both exist as x approaches a , then x → a f ( x ) ≤ lim lim x → a g ( x ) . As a consequence of this theorem, we get what is commonly known as the ”Squeeze Theorem”... Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 9 / 12

Section 1.4 The Squeeze Theorem If f ( x ) ≤ g ( x ) ≤ h ( x ) for all x in some open interval containing a (except for possibly at a itself), and x → a f ( x ) = L = lim lim x → a h ( x ) , then x → a g ( x ) = L . lim Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 10 / 12

Section 1.4 Here is a classic example of how to apply the Squeeze Theorem. � 1 Problem. Let g ( x ) = x 2 · sin � . What is x x → 0 g ( x )? lim (Why can’t we just use the product law for this problem?) Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 11 / 12

Section 1.4 Here is a classic example of how to apply the Squeeze Theorem. � 1 Problem. Let g ( x ) = x 2 · sin � . What is x x → 0 g ( x )? lim (Why can’t we just use the product law for this problem?) Solution. First we notice that the product law can’t be used because � 1 � lim x → 0 sin does not exist. (Why not?) x � 1 � However, since − 1 ≤ sin ≤ 1 for all x � = 0, we have that x � 1 − x 2 ≤ x 2 · sin � ≤ x 2 , and so the Squeeze Theorem implies that x x → 0 g ( x ) = 0 . lim Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 11 / 12

Section 1.4: Practice Problems 1 Let lim x → 2 f ( x ) = 4, lim x → 2 g ( x ) = − 2, and lim x → 2 h ( x ) = 0. Compute the following, or explain why the limit does not exist: (b) lim x → 2 [ g ( x )] 3 � (a) lim x → 2 [ f ( x ) + g ( x )] (c) lim x → 2 f ( x ) 3 · f ( x ) g ( x ) g ( x ) · h ( x ) (d) lim x → 2 (e) lim x → 2 (f) lim x → 2 g ( x ) h ( x ) f ( x ) 2 Let f ( x ) = x 2 + x − 6 and let g ( x ) = x − 2. Even though f ( x ) lim x → 2 g ( x ) = 0, lim x → 2 g ( x ) does exist. Find the limit (hint: try factoring f ( x )). 3 If 4 x − 9 ≤ f ( x ) ≤ x 2 − 4 x + 7 for x ≥ 0, find lim x → 4 f ( x ). (Use the Squeeze Theorem.) 4 Calculate the following limit by first simplifying the rational 1 / 4 + 1 / x expression: lim x →− 4 4 + x Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 26 August 2015 12 / 12