MATH 234, Calculus for BusinessLecture 5, Textbook Sections 3.1Introduction to LimitsUniversity of Illinois, Urbana-ChampaignFebruary 1st, 2017Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 1 / 25AnnouncementsHW2 is due Thursday 2/2, 11:59pm; late HWs are not accepted.Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 2 / 25Section 3.1, LimitsLimits give the prediction of the behavior of a function near a point,even if it is impossible to discuss the behavior at the point itself.ExampleLet f (x) =x2+ x − 6x2− 3x + 2. Note thatf (2) =(2)2+ (2) − 6(2)2− 3 ∗ (2) + 2=00which does not make sense (i.e., x = 2 is NOT in the domain of f (x)).Predict the behavior near x = 2.RemarkThere are two methods to predict behavior near a point/compute limits(1) Numerically/graphically(2) Algebraic Properties of LimitsFirst focus on Method 1Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 3 / 25Section 3.1, Limits Method 1: Make a table, draw a graphExample (... from previous slide)Let f (x) =x2+ x − 6x2− 3x + 2. Predict the behavior near x = 2.Solution (Numerical/graphical method)x approaches 2 from left x approaches 2 from rightx 1.9 1.99 1.999 2 2.001 2.01 2.1f (x) 5.444 5.040 5.004 undefined 4.996 4.960 4.636As x approaches 2 from left, then f (x) gets closer to 5.As x approaches 2 from right, then f (x) gets closer to 5.As x approaches 2 from both sides, then f (x) gets closer to 5.Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 4 / 25Section 3.1, Limits Method 1: Make a table, draw a graphExample (... from previous slide)Let f (x) =x2+ x − 6x2− 3x + 2. Near x = 2,x 1.9 1.99 1.999 2 2.001 2.01 2.1f (x) 5.444 5.040 5.004 undefined 4.996 4.960 4.636Definition (Limits from the left and the right)As x approaches 2 from left, then f (x) gets closer to 5: notation islimx→2−f (x) = 5which is read “ limit as x approaches 2 from the left of f(x) is 5”.As x approaches 2 from right, then f (x) gets closer to 5: notation islimx→2+f (x) = 5which is read “ limit as x approaches 2 from the right of f(x) is 5”.Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 5 / 25Section 3.1, Limits Method 1: Make a table, draw a graphExample (... from previous slide)Let f (x) =x2+ x − 6x2− 3x + 2. Near x = 2,x 1.9 1.99 1.999 2 2.001 2.01 2.1f (x) 5.444 5.040 5.004 undefined 4.996 4.960 4.636Definition (Two-sided limits, or, simply, limits)As x approaches 2 from both sides, then f (x) gets closer to 5: notation islimx→2f (x) = 5which is read “ limit as x approaches 2 of f(x) is 5”.RemarkFor a two-sided limit as x approaches a of f (x) to exist,Limits as x approaches a from left and right of f (x) must both exist.The left and right limits must give the same value.Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 6 / 25Section 3.1, Limits Method 1: Make a table, draw a graphExampleCompute limx→0−f (x), limx→0+f (x), and limx→0f (x), wheref (x) =|x|x=xxif x > 0undefined if x = 0−xxif x < 0SolutionNote that f (x) = 1 for x > 0 and f (x) = −1 for x < 0 andx -0.1 -0.01 -0.001 0 0.001 0.01 0.1f (x) -1 -1 -1 undefined 1 1 1So, limx→0−f (x) = −1 limx→0+f (x) = 1.The left and right limits exists, but do not give the same value. Solimx→0f (x) does not exist (abbr., DNE)Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 7 / 25Section 3.1, Limits Method 2: Use Algebraic Properties of LimitsTheorem (Algebraic Properties of Limits)Fix a and suppose limx→af (x) and limx→ag(x) both exist.(1) If k is constant, then limx→akf (x) = k limx→af (x).(2) limx→a(f (x) ± g (x)) = limx→af (x) ± limx→ag(x).(3) limx→a(f (x) ∗ g (x)) = limx→af (x) ∗ limx→ag(x).(4) limx→af (x)g(x)=limx→af (x)limx→ag(x), as long as limx→ag(x) 6= 0.(5) If p(x) is a polynomial, then limx→ap(x) = p(a).(6) If k is constant, then limx→af (x)k=limx→af (x)k.(7) If k is constant and positive, then limx→akf (x )= klimx→af (x).(8) If limx→af (x) > 0, then limx→alogb(f (x)) = logblimx→af (x)as long as b > 0 and b 6= 1.Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 8 / 25Section 3.1, Limits Method 2: Use Algebraic Properties of LimitsExample (Warm-up 1)Find limx→2f (x), where f (x) = x3+ 2x + 4.SolutionSince f (x) is a polynomial, property 5 giveslimx→2f (x) = f (2) = 23+ 2 ∗ 2 + 4 = 16Example (Warm-up 2)Find limx→6f (x), where f (x) = 24x.SolutionProperty 7 and 5 giveslimx→6f (x) = limx→624x= 2limx→64x= 24∗6= 224= 16,777,216Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 9 / 25Section 3.1, Limits Method 2: Use Algebraic Properties of LimitsExample (Warm-up 3)Find limx→3f (x), where f (x) = log4(253 + x).SolutionProperty 8 and 5 giveslimx→3f (x) = limx→3log4(253 + x) = log4limx→3253 + x= log4(256) = 4Example (Warm-up 4)Find limx→3f (x), where f (x) = log4(−253 + x).SolutionFor x near 3, then −253 + x < 0. The domain of log4is positive numbers.limx→3log4(−253 + x) does not existIntroduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 10 / 25Section 3.1, Limits Method 2: Use Algebraic Properties of LimitsExample (... the introductory example)Let f (x) =x2+ x − 6x2− 3x + 2. Compute limx→2f (x) using properties of limits.SolutionNote that limx→2(x2− 3x + 2) = 22− 3 ∗ 2 + 2 = 0 (denominator goes to 0).However,limx→2x2+ x − 6x2− 3x + 2= limx→2(x − 2)(x + 3)(x − 2)(x − 1)Factor= limx→2x + 3x − 1Cancel=limx→2x + 3limx→2x − 1Property 4=2 + 32 − 1= 5 Property 5Introduction to Limits (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 1st, 2017 11 / 25Section 3.1, Limits (Non-)Existence of LimitsExample (Ways for limits to not exist)Fix the number a. The expression limx→af (x) does not exist iflimx→a−f (x) 6= limx→a+f (x) or if either directional limit is undefined.limx→a−f (x) = ±∞