Broward College - MAC 2311Calculus Exam 1 ( all solutions are 100% correct )

Document Content and Description Below

MAC 2311 Learning Activity #4 Special Limits; Continuity Name: Group #: Evaluate the following limits using appropriate methods: 1. lim x!0+ tan"1 (ln x) 2. lim x!1 1 ! ex 1 + ex 3. lim x... !0 3 cos(x) sin(4x) 2x 4. Given there exists a function h(x) such that 2x3 + 5  h(x)  x4 + 5 for all x near 2, find lim x!2 h(x) + 3 4 Page 1 of 3 Apply HMHOHAMWK glxttncxlflxtarctanx) llmlaktanllhlxl) x -70T "Mxtxlaratnlxl) llmxtotllnlx)) -- -X = - M2 amaemex.ie#ieiIexIiitfIIiimmx::IteiiexI , "M×→pl4ex -D= -1 "M×→wl' text) 't ⇒ -41=-1--3/2 .HM×→o( 108k¥44) ⇒ 312 .hn#of-slhXSM4Xt4Los=3l2tlMx-sotSlnxsln4Xt4COS4xcosxjl4HlfKH ) plugin 0=3/21-Slnlolslhlllo) -410514010101=3/214-01 4. 3/2--3.4/2=12/2--6 sandwich Theorem ""4JENHfLEx4tq#k→z2xfI 'MxhdhK¥3=f 1×3-18 EACH -13£ y4+g '2234-+8=29/9--6 4- -4 IT ttxtz 2941=6MAC 2311 Learning Activity #4 Special Limits; Continuity 5. Using the conditions of continuity, determine if f(x) is continuous at x = 0 and x = 1. If f(x) is discontinuous at either point, indicate the type of discontinuity. f(x) = 8<: 1 ! x : x < 0 ex : 0  x  1 x2 + 2 : x > 1 6. Find a value of k for which g(x) is continuous everywhere: g(x) = 8<: x2 ! 9 x ! 3 : x 6= 3 kx : x = 3 Page 2 of 3 HMxyo-flxt-llmy-otftxtllmxso-lt-XI-llmy-otle91-O.CO 1=1 function ntmoosatxo HI ⇒llmx-y-lt-llmy.lt (112-12) e' =/ -12 ⇒ f-3 Oswald .FM/7MdlscohhnwsAtX--lib0thSlAlllMHlXlSt jump discontinuity Hmysgglxl -913) x -73,1143 "Myers I=3k 3-13=312 "Mx→gX2 -32 6=312 ⇒ - - 3K 4=2 " my-zH¥¥t3=3kMAC 2311 Learning Activity #4 Special Limits; Continuity 7. Consider the function f(x) that satisfies the following conditions: • f(x) is continuous for all x except x = !4, x = 0, and x = 4. • f(0) is undefined • lim x!0 f(x) = 3 • lim x!"4! f(x) = !1 lim x!"4+ f(x) = 1 • lim x!4! f(x) = 1 lim x!4+ f(x) = !1 • lim x!"1 f(x) = 2 lim x!1 f(x) = 2 (a) State the equation(s) for any horizontal asymptote(s) on the graph f(x). (b) State the equation(s) for any vertical asymptote(s) on the graph f(x). (c) At what x-value(s) does f(x) have any discontinuities? What type are they? If there are none, state that. (d) Sketch a possible graph of f(x). Page 3 of 3 "Mys -yfall --2 & Ilm x →WAX) --2 SO hormonal asymptote IS y --2 "tf -y - f-IN = -X & 11 My -7 -ytflx) =D and vertical asymptote X -- -4 & X -- 4 11MW4- fall =p tell my-741- fall =-D at x -- -4,11=0 And A-4 Is discontinues → at x= -4 ; Hmx ,-4 - FX#ftp.ytx non- removable discontinuity of Infinite type → at X --o ; Amy, ofHHS & flownde . #"""÷÷÷÷÷÷÷÷÷÷÷÷÷:#MAC 2311 Learning Activity #0 Precalculus Review Name: 1. Let f(x) = 3x2 ! x. Find the following: (a) f(1) (b) f(!4) (c) f(a + 1) 2. State the domain and range of the following functions: (a) f(x) = x2 + 1 (b) f(x) = px + 3 (c) f(x) = 9 x ! 2 Page 1 of 6 Valentina Espinosa 3111411--2 31-414-41=52 3 ( Atl)' - ( Atl) -- 31942Atl) - - Ia ' + bats -lati) = ya't Sa -12 DOMAIN : I -0,0) Range :[tix) Domain :C-3,0) Range : coin) DOMAIN : l - O , 2) V12 , D) Range : I - o , ow loin)MAC 2311 Learning Activity #0 Precalculus Review 3. Use the function given to evaluate the following: h(x) = 8<: 1 ! x5 : x < 3 ex : x = 3 8 : x > 3 (a) h(5) (b) h(!1) (c) h(3) 4. Let f(x) = px2 ! 1 and g(x) = x + 1. Find the following: (a) ✓fg ◆ (x) (b) f(g(x)) (c) (g # f)(2) Page 2 of 6 ht) : X > 3 : 8 ht - l ) : X Lg : I - XS : I - f- 1) 5 : 2h13 ) : X =3 : ex : ( E3) : 20.8553 XI RE Txt : expand Http - l : X2t2X TEX ( Xt 1)(F)(2) : al btc ) -- ab -Iac gifted) -- gas -- tf -itMAC 2311 Learning Activity #0 Precalculus Review 5. Find the equation of the line through the point (1, 7) with slope 2 3 . Present your answer in slope-intercept form. 6. Find the solutions to the following: (a) 1 ! (4 ! (5 ! x)) = 3x ! (4 + x) (b) x2 + 5x = !4 (c) 2✓ cos(✓) + ✓ = 0 on [0, 2⇡] Page 3 of 6 f- Mxtb 7=21311 )tb=b= 191g 4=213-11913 -X-12=2×-4 -X -24=2×-6 -2X -Y-12-2=2×-4-2 -3x=-6 -4=2×-6 4=2 "I '¥÷¥¥¥o " -b X -- -5fF 2. I 201010=-0 101--42 coslhtnbtecosocospnk - %) 1010=044%1 coso.ca/4nls/ 0--10,243,4%1MAC 2311 Learning Activity #0 Precalculus Review 7. Given sin(✓) = 12 13 and ⇡ 2 < ✓ < ⇡, find the other 5 trigonometric values. 8. Find the inverse, f !1(x), of the following functions. (a) f(x) = 4x ! 1 2x + 3 (b) f(x) = ln(5x ! 1) Page 4 of 6 2 SINO -12113 - X -- -Trey 2 Colo HMO---- -His 145 -CSOO --13/12 -RIP Seco= -Bst -off = -5 Coto-542 " hh xkytst.IT#l2ytH=xl2ytD--4y-IY---3x-I 24-4 Inky - D l " --Ty -I Y --e¥=Y=t¥MAC 2311 Learning Activity #0 Precalculus Review 9. Solve the following equations: (a) 42x!3 = 64 (b) 4e2x ! 7ex = 15 (c) log3 x + log3(x ! 6) = 3 10. Using the properties of logarithms, express the given quantity as a single logarithm: ln(a + b) + ln(a ! b) ! 2 ln c Page 5 of 6 42×-9=43 2X -3=3 X -- 3 LIV '-70=15 2 402-745=0 -"tr? U --3 , A- -514 logglxlx -A) =3 XIX- 61=27 xlx-bt.no' tfaifelnla-ibka-bl2lnclnla-lbka-bt.HR MattyMAC 2311 Learning Activity #0 Precalculus Review 11. Evaluate the following. Keep in mind the restricted function values for inverse trig functions. (a) cos!1 p2 2 ! (b) arctan(p3) 12. Determine if the following functions are even, odd, or neither. (a) f(x) = 5x4 ! 3x2 + 7 (b) f(x) = 3x ! 6 (c) f(x) = 4x5 + 2x3 ! x Page 6 of 6 are KEI -- 44 "B ft-X)-174-3×47 -HY) -- - 544+9×2-7 f- ( X ) --ffx) f- IX) # fl -X ) 5×4-3447 EVEN not Odd A- x) : -3×-63×-6 Atx) -- -3×-6 -th) ---3*6 -fly): -Htb neither f-txt. -4×5-2×44 -Htt. -4×5-2×3.tx AHHHH) - Hxtffx) hotenon OddInfinite Limits & Limits at Infinity MAC 2311 Florida International University 1 Infinite Limits An infinite limit occurs when as the independent variable approaches a value, the dependent variable becomes arbitrarily large in magnitude. So, the function values increase or decrease without bound near a point. Infinite limits are denoted by: lim x!a f(x) = 1 or lim x!a f(x) = "1 In the case of one-sided infinite limits: lim x!a! f(x) = 1 or lim x!a! f(x) = "1 lim x!a+ f(x) = 1 or lim x!a+ f(x) = "1 Note: In the event of an infinite limit, the limit does not exist since as x approaches some number, the function values become increasingly large in magnitude, never approaching a single, unique value. However, we don’t use the notation of “DNE” to express this since we have a formal way to describe this occurrence. Example 1: Evaluate the following limits using the graph of g(x) given below. • lim x!2! g(x) • lim x!2+ g(x) • lim x!2 g(x) • lim x!4! g(x) • lim x!4+ g(x) • lim x!4 g(x) 1 W ONE - X . -X -X -y1.1 Finding Infinite Limits Analytically Many infinite limits are analyzed using a common arithmetic property: The fraction a b grows arbitrarily large in magnitude if the denominator, b, approaches 0 while the numerator, a, remains nonzero and relatively constant. In other words, lim b!0 a b = ±1. Let’s take the following limits for example: lim x!0+ 1 x lim x!0! 1 x lim x!0 1 x In all three cases, direct substitution does not work because we would be dividing a nonzero number by zero. So, let’s observe what happens as we let x approach zero from either side. In the tables below, we can approach this numerically and by using the graph of f(x) = 1 x , we can observe what happens graphically. x approaches 0 from the right: x 1 0.1 0.01 0.001 0.0001 0.00001 1x 1 10 100 1, 000 10, 000 100, 000 x approaches 0 from the left: x "1 "0.1 "0.01 "0.001 "0.0001 "0.00001 1x "1 "10 "100 "1, 000 "10, 000 "100, 000 We see, as x approaches zero from the right and left, the function values grow larger in magnitude. Specifically, as x approaches zero from the right, the function values grow more and more positive, while as x approaches zero from the left, the function values grow more and more negative. So, we can draw the following conclusions for each limit: lim x!0+ 1 x = 1 lim x!0! 1 x = "1 lim x!0 1 x DNE 2Let’s look at the left-hand limit ✓xlim !0! x1◆ and solve it analytically (by figuring out how the function behaves without actually plugging in several values or without having to look at its graph). • Since x is approaching zero from the left, this means that the x-values will be less than zero. So, they will be negative. • In the numerator, what x is approaching does not matter since 1 is a constant. [Recall the first of our common limits in techniques for computing limits.] • For the denominator, x is getting closer to zero and is negative. • So, overall, we have a positive constant being divided by an increasingly small negative number. The result of this fraction will be an increasingly large negative number. This gives us the following: lim x!0! approaches 1 z}|{ 1 x |{z} approaches 0, negative = "1 Example 2: Evaluate each of the following limits: • lim x!1+ !"2 z}|{ "2 x " 1 | {z } !0,+ = "1 Since we are approaching 1 from the right, this means that x > 1. This implies x " 1 > 0. So, x " 1 will get closer to zero and will be positive as x ! 1+. Therefore, we have a negative constant being divided by an increasingly small positive number. The result will be a number which gets larger in magnitude and negative. So, the limit will be "1. • lim x!1! "2 x " 1 • lim x!2! x + 1 (x " 2)2 3 - -2 * - - X -- I vertical asymptote approaching I from HA X Ll → x -1=0 = D - s tot -- Nth¥. ttmztxth me . thx-255-4=2 Vertical asymptote x' ' Mz . ( Xt 1) =3 ¥72 - ( ' H -2)2) =p b - b =D1.2 Vertical Asymptotes For any given function f(x) if any of the following limits are true, then f(x) will have a vertical asymptote at x = a. lim x!a! f(x) = ±1 lim x!a+ f(x) = ±1 lim x!a f(x) = ±1 Only one of the above limits has to occur in order for a function to have a vertical asymptote at x = a. Example: Let’s revisit the function f(x) = 1 x . In previous work, we found the following: lim x!0+ 1 x = 1 and lim x!0! 1 x = "1 From this we are able to draw the conclusion that the function f(x) = 1 x has a vertical asymptote at x = 0. Finding Vertical Asymptotes of Rational Functions Given a rational function, p(x) q(x), we can determine the vertical asymptotes, if any, with the following steps. • Find the x-values where the denominator is zero, but the numerator is NOT zero. – Find x = a such that q(a) = 0, but p(a) 6= 0 • Using the value(s) identified in the previous step, take one-sided limits of the function. – Find lim x!a+ p(x) q(x) and/or lim x!a! pq((xx)) • If either of the above limits go to ±1, then your function has a vertical asymptote at x = a. Example 3: Identify any vertical asymptotes of f(x) = x2 + x " 2 x2 " 4x + 3 4 HYHX -12) ( X -HNK) -⇒Fz, = Xt2/X-3 112-4×-13 Compareto 20N x-5-0 * I 9111=0 putt Kim 4--3*99131--000140 , ¥3 --z HY HA-HA I qq.fi#ymx*yzHAs-EE-- * µm→I tone2 Limits at Infinity A limit at infinity occurs when the independent variable increases or decreases without bound. So, limits at infinity tell us how a function is behaving as its x-values get increasingly more positive or negative. Limits at infinity are denoted by lim x!1 f(x) = L or lim x!"1 f(x) = L Example: Evaluate lim x!1 1 x and lim x!"1 1 x We note that direct substitution is not a valid approach for evaluating these limits since we can’t simply plug in the values of ±1. So, there are a couple of other approaches we can take. • Graphically: Let’s look at the graph of f(x) = 1 x and observe what happens as the x-values grow larger in magnitude. From the graph, we see that as x increases and decreases more and more, the function values get smaller and smaller in magnitude, approaching the value of zero. So, we are able to conclude: lim x!1 1 x = 0 and lim x!"1 1 x = 0 • Analytically: Similar to analyzing infinite limits, we can analyze limits at infinity using an arithmetic property: The fraction a b grows arbitrarily small in magnitude if the denominator, b, grows arbitrarily large in magnitude while the numerator, a, remains nonzero and relatively constant. In other words, lim b!±1 a b = 0. For f(x) = 1 x , as x ! 1 the numerator stays constant at 1 and the denominator grows larger in magnitude and positive. Therefore we have a positive constant being divided by an increasingly large positive number. The result of this fraction will be an increasingly small, positive number. Therefore, lim x!1 1 x = 0. 5Now, as x ! "1, again the numerator stays constant at 1, but the denominator grows larger in magnitude and negative. So, we have a positive constant being divided by an increasingly large negative number. The result of this fraction will be an increasingly small, negative number. This means we have lim x!"1 1 x = 0. 2.1 Horizontal Asymptotes For any given function f(x) if either of the following limits exists, then f(x) will have a horizontal asymptote at y = L. lim x!1 f(x) = L or lim x!"1 f(x) = L Note: A horizontal asymptote can only occur when the limit at infinity yields a finite number. Example: Let’s again revisit the function of f(x) = 1 x . We just found that lim x!1 1 x = 0 and lim x!"1 1 x = 0. From this we are able to conclude that f(x) = 1 x has a horizontal asymptote at y = 0. Example 4: Find the horizontal asymptotes of the following functions. [Hint: Consider the limit laws.] • f(x) = "2 x • f(x) = 3 x + 5 2.2 Infinite Limits at Infinity If a function f(x) increases or decreases without bound as x increases or decreases without bound, then we have an infinite limit at infinity. These are denoted by: lim x!1 f(x) = ±1 or lim x!"1 f(x) = ±1 Here are some examples of infinite limits at infinity: • lim x!1 x = 1 • lim x!"1 x2 = 1 • lim x!1 ex = 1 • lim x!"1 x3 = "1 Note: In the case of infinite limits at infinity there exists no horizontal asymptotes. 6 -2/4=4--0 x' 'Boitano horizontal asymptote -- O ' x'F.ohhh 496134-15 nggx-iygg.es, Y=MXtb '- 0-15=5 -' horizontal y=5 Y horkohtalasympt . -52.3 Limits at Infinity of Powers and Polynomials Let n be a positive integer and let p be the polynomial p(x) = anxn + an"1xn"1 + · · · + a2x2 + a1x1 + a0 where an 6= 0. • If n is even: lim x!±1 xn = 1 • If n is odd: lim x!1 xn = 1 and lim x!"1 xn = "1 • lim x!±1 1 nx = 0 • lim x!±1 p(x) = 1 or "1, depending on the degree (odd or even?) of the polynomial and the sign of the leading coe!cient an (positive or negative). – lim x!±1 p(x) = lim x!±1 anxn = an · lim x!±1 xn Example 5: Evaluate the following: • lim x!"1 (7x5 " 4x3 + 2x " 9) = lim x!"1 7x5 = 7 · lim x!"1 x5 = 7 · "1 = "1 • lim x!1 (8x2 + 3x " 5x3) • lim x!"1 (17x3 " 4x9 " 5x + 1) • lim x!"1 "2 x3 • lim x!1 18 7 . Infinite property My ( axht . . .bXtC) -- HALO 414655×9=-5×191×2-51 # =p A- -5 As = - Y Infinite property :a= -4 A- a =p fifty-4k¥ -411M¥ , ?Lolxal= -41-07-0 "May ( 4×4=0 Hmx sac -- C = 182.4 Limits at Infinity for Rational Functions In general, for rational functions, we can evaluate limits at infinity by dividing each term in the expression by the highest power of x in the denominator, simplifying, and then evaluating the limit (for each term). Example: Evaluate lim x!1 3x + 2 x2 " 1 Since this is a limit at infinity, direct substitution is not a valid approach. However, given this is a rational function, we can divide each term by the highest power in the denominator. For this function, the highest power in the denominator is x2. So, we have the following: lim x!1 3x + 2 x2 " 1 = lim x!1 3x x2 + x22 x2 x2 " 12x = lim x!1 3x + 2 x2 1 " 1 x2 = lim x!1 3 x + lim x!1 2 2x lim x!1 1 " lim x!1 1 2x = 0 + 0 1 " 0 = 0 For the example, since we found that a limit at infinity led to a finite value, we are able to say f(x) = 3x + 2 x2 " 1 has a horizontal asymptote at y = 0. Example 6: Evaluate lim x!1 x + x3 " 8x4 2x4 + x2 " 1 Example 7: Evaluate lim x!"1 x5 + 4x2 " 5 x3 + 6x 8 - XIX ' - 844 x"Mg # t - 8¥44 Tix ÷ dmadbya,Y4 Html-# tf - 8=-8 I 2-14×2-11×4=2 4444¥ -¥1 all tntgfff -812=-4 dlndlbyhlghestdenpouer -- X ' 44M¥55 -14¥ divide all sections Dyxs -- retire -- X ' -1¥-¥ -- x - - ¥i¥ ¥ -- y l -1¥ = I qq.nu#*m.ofxI-5tiF-oYxs - ' 47 t MalkySpecial Limits & Continuity MAC 2311 Florida International University 1 Special Limits Here we present some limits that will be helpful in evaluating other limits. You’ll want to keep these in mind as you work through certain types of limit problems. • lim x!1 ex = 1 • lim x!#1 ex = 0 • lim x!1 ln(x) = 1 • lim x!0+ ln(x) = "1 • lim x!1 tan#1(x) = ⇡ 2 • lim x!#1 tan#1(x) = "⇡ 2 • lim x!0 sin(x) x = 1 • lim x!0 1 " cos(x) x = 0 Example: Evaluate the following: lim x!0! e 1x Since we cannot divide by zero, direct substitution will not give a valid output. However, we can attempt to simplify this problem by rewriting the expression so that it resembles one of the special limits we have. The “complex” part of the expression is the exponent, 1 x. Let’s use a change of variable in order to simplify this piece and rewrite our limit in terms of the new variable. Let t = 1 x . With this change of variable, we can rewrite the function we are taking the limit of to be in terms of t: e 1x = et Now, we need to write the value the independent variable, x, is approaching in terms of our new variable, t. In our original limit, x is approaching 0 from the left. So, what does this mean for our new variable, t? lim x!0! t = lim x!0! 1 x = "1 Using an infinite limit, we see that as x ! 0#, t ! "1. We can now rewrite our limit in terms of t and use one of our special limits to evaluate: lim x!0! e 1x = lim t!#1 et = 0 1Example 1: Evaluate the following: lim x!2+ tan#1 ✓x "1 2◆ Example 2: Evaluate the following: lim x!0 sin(5x) x Example 3: Evaluate the following: lim x!0 2 sin(x) " cos(x) + 1 x [Hint: This problem does not require a change of variable. Instead, try using algebra to rewrite the expression.] 2 x'the ' to 'ftp.tf#IIo=sxnt.tstx ftp.fantxt.zt-x.tzt-fjnftan-ttt.tk 417055M¥ " Ifm! Ingenito ⇒ not -0 . limo 'Yt¥f5sn¥ -- Etim sifts Timok'hfI -' HII ) 'ftp.nf#tflMotofI--2.ltO-- I1.1 The Squeeze Theorem The Squeeze Theorem If the functions f(x), g(x), and h(x) satisfy the following for all values of x near a: f(x)  g(x)  h(x) and if lim x!a f(x) = L = lim x!a h(x), then lim x!a g(x) = L. Note: The value a may be a finite number or ±1. A graphical example of the Squeeze Theorem is given below. Example: Evaluate the following: lim x!1 sin(x) x Using the methods for evaluating limits we’ve covered so far, we don’t have a way to solve this limit. So, let’s try to approach this using the Squeeze Theorem. First, we note the following by using the known range of the sine function. "1  sin(x)  1 Since x ! 1, it is reasonable to assume that x > 0. So, using properties of inequalities, "1 x  sin(x) x  1 x Taking limits at infinity of the two outer functions we get: lim x!1 "1 x = 0 and lim x!1 1 x = 0 Therefore, by the Squeeze Theorem, we are able to conclude lim x!1 sin(x) x = 0. 3Example 4: Given there exists a function, f(x), such that 2x " 1  f(x)  x2 for all x, find lim x!1 f(x) " 1 x " 1 . 2 Continuity The graph of a function can be described as a “continuous curve” if it is “unbroken”. In this section, we will explain the idea of an “unbroken curve” by using a mathematical property known as continuity. To make this idea more precise we need to develop some properties that make a curve “unbroken” or continuous. 2.1 Conditions for Continuity A function f(x) is continuous at a point x = a if each of the following conditions are upheld: 1. f(a) is defined 2. lim x!a f(x) exists 3. lim x!a f(x) = f(a) If one or more of these conditions are not upheld, then f(x) is not continuous at x = a and x = a is a point of discontinuity. Example 5: Using the checklist for continuity, determine if f(x) is continuous at x = 3. f(x) = 8<: x2 : x < 3 x + 6 : x > 3 10 : x = 3 4 j - Ix - HEHN -11×2-1 ' x'Fi --2Y solo XT XT XT 41mg .tk#I2X-2LfNl.lLXH4llMl2I=2 It Ii XT tht tf # 4hm time '' solo HFit¥=2 HANH ) -- HI -2 I. HH -- IOV 3¥MzHHtH3l ' ftp./xy=czp..qfHllsnotantnovsx=sHFhHHIffsh+HtH-- 'f'F-HH - stag-7HAzHH=aV2.2 Types of Discontinuity If a function f(x) is not continuous at a number a, we say f(x) has a discontinuity at x = a. • Removable discontinuity (hole): lim x!a! f(x) = lim x!a+ f(x), but lim x!a f(x) 6= f(a) • Jump Discontinuity: The function “jumps” from one value to another at the point x = a. lim x!a! f(x) 6= lim x!a+ f(x) • Infinite Discontinuity: There is a vertical asymptote at x = a. lim x!a f(x) = ±1 or lim x!a! f(x) = ±1 or lim x!a+ f(x) = ±1 5To determine where a function is discontinuous, use similar rules for determining the domain of that function (we can’t divide by zero, can’t take the even root of a negative number, etc.). Then, if needed, use limits to determine the type(s) of discontinuity that occurs at these points. Example 6: Determine if the following functions have points of discontinuity. If so, determine what type each are. • f(x) = 3 4 " x • g(x) = ⇢ x 2x2 ""9 : 5 : xx >  3 3 • h(x) = x2 " 25 x " 5 6 4¥10 IIF ' ' 't ''The.iq?=-x that"¥.tn#.I...wFmxfI'" one HXINUSANINHNH dlklhtnutyatx.LI HMGlH=HMH2 -a) llmglxtthmglxl 1173- X-23 - ' - (32-9)=0 X-D - Xtzt 11-13 HMglH=HMl2X KB -5--2131-5=1 guinasajumpdlskh . Atx'S X - 5=0 HMCHXHHM 11=5 XT Xt ' "m - -0/0 ¥E;xts 44ms - " ⇒ 4,71*51=51570 15*5 HMh¥=IO HMNXHO X -75 hlklhasatemovabledlscohtinutyatx -- s2.3 Continuity Rules If two functions, f(x) and g(x), are continuous at a point x = a, then the following are also continuous at x = a: • f(x) ± g(x) • cf(x) • f(x) · g(x) • f(x) g(x), as long as g(a) 6= 0 • f(g(x)), as long as f is continuous at g(a) • [f(x)]n, where n is a positive integer These rules are important because we can use them to find that: • Polynomial functions, p(x), are continuous everywhere. • Rational functions, p(x) q(x), are continuous everywhere EXCEPT where q(x) = 0. • A function [f(x)]n/m with m odd is continuous at all points where f(x) is continuous (m, n are positive integers with no common factors). • A function [f(x)]n/m with m even is continuous at all points that f is continuous and f(a) & 0 (m, n are positive integers with no common factors). • The trigonometric functions sin(x) and cos(x) are continuous everywhere, while the functions tan(x), sec(x), csc(x) and cot(x) are continuous at every point in their domain (remember that their domains are restricted). 7MAC 2311 Learning Activity #3 Infinite Limits; Limits at Infinity Name: Group #: 1. Consider the graph of f(x) in the graph below. (a) Find the following: i. lim x!"4! f(x) ii. lim x!"4+ f(x) iii. lim x!1 f(x) iv. lim x!3+ f(x) v. lim x!3! f(x) vi. lim x!"1 f(x) (b) State any horizontal asymptotes for f(x). If there are none, state that. (c) State any vertical asymptotes for f(x). If there are none, state that. Page 1 of 3 =D = - y = 6 =D = -2 =/ 4=-2 4=1 X= - 4 X =3MAC 2311 Learning Activity #3 Infinite Limits; Limits at Infinity 2. Evaluate the following analytically: (a) lim x!8+ !5 x ! 8 (b) lim x!"1 (7x7 ! 4x3 + 2x ! 9) (c) lim x!3! 4 3 ! x (d) lim x!0+ 1 x + 3 (e) lim x!"4! 3x x + 4 (f) lim x!"1 (1 ! 2x + 5x2 ! 17x3 ! 4x8) (g) lim x!"5+ (x + 5)2 x2 + 5x (h) lim ✓!⇡! cot(✓) 3. Find any horizontal and vertical asymptotes of f(x) = 3x2 x ! x2. If there are none, state that. Page 2 of 3 B. llmxt -4 - ( XIX-14 ) Xapppoaohngsrfromrlght and 'ht4xxq¥' Item, 1158 X-870 -- - D 'En HMX , # Calm -1 . .bXtC) Infinity property = - HALO a > o a- 7h --7 =-D A-even A--4nF -y Xapppoaomgsofromleftylz factor "" xryx plug FEI) y - X ) O = y =o plvgo ftp.llgcomnahlmttllmfsn-kott) = - X simplify 34/14 simplify 3M - XH -- SHH factory -x2VXltH I - X -- O X - I 3×4×11 -X) vertical lymph 4=3/-1 4=-3 X -X?- 0=7×11 -X)=O=1X=Qx=I hohlohtalasymp . -3MAC 2311 Learning Activity #3 Infinite Limits; Limits at Infinity Evaluate the following limits and give the equation for any horizontal asymptote (if there are none, state that). 4. lim x!1 !2x3 ! 2x + 3 3x3 + 3x2 ! 5x 5. lim x!"1 4x3 + 6x2 ! 2 5 ! 4x + 2x2 6. lim x!1 p3 x + 4 1 + 7px Page 3 of 3 dlvldlhkyhlttaenom . X's leading coefficient and the refine numerator. -2 -2 -4×2+4×3 denominator :3 .syµ=I2 = -43 horizontal -213 ymp . dMAlhlghlttdlno.lk And refine 4YX2= - y -92=-0 51×2-41×-12=2 dlvkllallbyrx retire + 1-4- =D bfx TX 017=0 1×5+7= 7An Introduction to Limits MAC 2311 Florida International University The concept of a “limit” is the fundamental building block on which all calculus concepts are based. This section is intended to present an informal view of limits with the ultimate goal of developing an intuitive understanding for the basic ideas. 1 An Intuitive Approach to Limits 1.1 Rates of Change One of the most important characteristics of functions is how a function behaves or changes over time. For example, we may be interested in how much the value of a function changes over a certain interval. This leads to a quantity known as the average rate of change of the function. Given any function, y = f(x), we find the average rate of change of y with respect to x over an interval from x1 to x2 by taking the ratio of the change in y, !y = y2 ! y1 = f(x2) ! f(x1) and the change in x, !x = x2 ! x1. So, the average rate of change of a function y = f(x) over an interval [x1, x2] is found by: Average rate of change = !y !x = f(x2) ! f(x1) x2 ! x1 = y2 ! y1 x2 ! x1 Example 1: Find the average rate of change of f(x) = sin(x) over the interval h0, ⇡6 i. 1 ARV -- FI = SIN M¥110) Mb - O = ' knife → tf ' -- th . bln = In -- H nExample 2: A rock is launched vertically upward from the ground with a speed of 85 ft/sec. Not considering air resistance, the position of the rock after t seconds is given by s(t) = !16t2 + 85t. find the average rate of change of the rock between 1 and 3 seconds. Let’s say we have a function f(x) that contains two points, P: (x1, y1) and Q: (x2, y2), on its curve. Geometrically, the slope of the line between two points on a curve gives us the average rate of change between the two points. A line through two points on a curve is called a secant line. The slope of a secant line is the average rate of change between the two points. A line that just touches the curve at a point and has the same direction of the curve at that point is called a tangent line. Given the geometry of a tangent line to a curve at a single point, it is feasible to identify the slope of the tangent line at a point as the slope of the curve at that point. The slope of a tangent line is the instantaneous rate of change at a single point. So, the average rate of change gives the rate of change of a function across an interval while the instantaneous rate of change gives the rate of change of a function at one particular point. 2 SHH -HAI -18ft A- SCH - SH ) -16×4+85 Ft =2IfHSlC (-16×32+85237176×12184) 21.2 Connecting Average and Instantaneous Rates of Change Finding a Tangent Line To find the tangent line at a point P: • Imagine there are two points on a curve, P and Q. • Let the point Q move closer to the point P along the curve. • As Q moves closer to P, the secant lines between P and Q begin to approach the tangent line at P. • So, as Q gets closer to P, the slopes of the secant lines between P and Q approaches the slope of the tangent line at P (As Q ! P, mP Q ! mP) The slope of the tangent line is the limit of the slopes of the secant lines as one point approaches the other. Finding Instantaneous Rate of Change Let’s again consider how we find the average rate of change between two points on a curve. If we compute the average rate of change of y = f(x) over the interval [x0, x0 + h], then the change in x is represented by h. In other words, !x = (x0 + h)!x0 = h. So, the average rate of change is calculated as !y !x = f(x0 + h) ! f(x0) h . To find the instantaneous rate of change at x0, the change in x would have to be zero in the average rate of change. However, we see that we cannot substitute in h = 0 since we cannot divide by zero. So, if we want to find the instantaneous rate of change of a function f(x) at a point x = x0, • Compute the average rates of change over intervals that get shorter and shorter (approaching the point of interest, x0). • As the length of intervals decrease, the average rates of change begin to approach a unique number. • This unique number is the instantaneous rate of change at x0. The instantaneous rate of change is the value the average rate of change approaches as h, the change between the two points, approaches zero. 3Example 3: Determine the instantaneous rate of change of f(x) = x2 at x = 3. • Find the average rates of change over the given time intervals. Do not round your intermediate calculations. – [3, 4] – [3, 3.5] – [3, 3.1] – [3, 3.01] – [3, 3.001] • Estimate the instantaneous rate of change when x = 3. 4 HAI b-a ftp.f#Hfs7 ftp.ftjf# → ' → b 's Afl ) - f-(3) ⇒ → 9fl → bi tf 3.011 -fl'd 9.0601-9 → → 6.01 H3j%H → 909881L → b.ooi-nhnfflsth.tt# → Bth h IROC →Ath2ntbh → HY → b2 The Limit of a Function The most basic use of limits is to describe how a function behaves as the independent variable approaches a given value. For example, let’s examine the behavior of the function f(x) = x2 ! x + 1 as x-values get closer and closer to 2. We can see from both the graph and the table of function values, as x approaches 2 from the left and right sides, the function values approach 3. We would describe this by saying “the limit of x2 ! x + 1 as x approaches 2 is 3”. Limit Notation: For a function f(x), as x gets arbitrarily close to a, if f(x) gets arbitrarily close to L, then we say that “the limit of f(x) as x approaches a is L”. We use the following notation to express this: lim x!a f(x) = L It is important to note that limits only tell us how a function is behaving near a point and not what is happening at the point. Take the following for example: We notice that in each of the graphs illustrated, the limit of f(x) as x approaches a is 2, or lim x!a f(x) = 2. However, in the far left graph f(a) is undefined and in the middle graph f(a) = 3. Again, the limit of f(x) as x gets arbitrarily closer to a only tells us how the function is behaving near a, not what is actually happening at x = a. 5Example 4: Evaluate the following limits using the graph of f(x) given below. • lim x!"1 f(x) = • lim x!1 f(x) = • lim x!2 f(x) = 2.1 One-Sided and Two-Sided Limits The limit, lim x!a f(x) = L, is referred to as a two-sided limit because it describes what happens as a function approaches an x value from both sides, left and right. However, some functions exhibit behaviors that di↵er on either side of an x-value. In this case, it is necessary to indicate whether x-values near a point are on the left side or right side of the point in order to determine limiting behavior. One-Sided Limits For a function f(x), as x gets arbitrarily close to a, but always less than a, if f(x) gets arbitrarily close to L1, then we say “the limit of f(x) as x approaches a from the left is L1”. We use the following notation to express this left-hand limit: lim x!a! f(x) = L1 For a function f(x), as x gets arbitrarily close to a, but always greater than a, if f(x) gets arbitrarily close to L2, then we say “the limit of f(x) as x approaches a from the right is L2”. We use the following notation to express this right-hand limit: lim x!a+ f(x) = L2 Note: The superscript of “!” indicates a limit from the left of x = a and a superscript of “+” indicates a limit from the right of x = a. 6 2 - I 0Relating One-Sided and Two-Sided Limits The two-sided limit of a function f(x) exists at x = a if and only if both of the one-sided limits exist at a and have the same value. Therefore, if the right- and left-hand limits do not agree (are not the same value), the two-sided limit fails to exist. That is, lim x!a f(x) = L IF AND ONLY IF lim x!a! f(x) = L = lim x!a+ f(x) For example, consider the function f(x) = |x| x . From the graph of f(x) (see below), we are able to see that lim x!0! |x| x = !1, while lim x!0+ |x| x = 1. Since the left- and right-hand limits do not agree, the two-sided limit at x = 0 will fail to exist for f(x) = |x| x . In general, when the limit of a function fails to exist at a point, we use the expression of “DNE” to indicate that it “does not exist”. So, for the above example, we would say: lim x!0 |x| x DNE. When Limits Fail to Exist • If lim x!a! f(x) = L1 and lim x!a+ f(x) = L2, where L1 6= L2, then lim x!a f(x) fails to exist. – If lim x!a! f(x) 6= lim x!a+ f(x), then lim x!a f(x) DNE. • If the function oscillates too much around a point, then the limit of the function at that point will not exist. – For example, consider the function f(x) = sin ✓x1◆ 7x 0.01 0.001 0.0001 0.00001 0.000001 sin % x1& !0.5064 0.8269 !0.3056 0.0357 !0.3450 From the graph of f(x), it’s di#cult to tell what is happening near x = 0. So, let’s look at a table of function values of f(x) at x-values getting closer to 0. We see that as x ! 0+, the function values “jump around” never approaching one unique number. Because of this, the limit as x approaches 0 fails to exist. Or, lim x!0 sin ✓x1 ◆ DNE. Example 5: Use the graph of f(x) below to determine the following values, if they exist. • f(1) • lim x!1! f(x) • lim x!1+ f(x) • lim x!1 f(x) • f(2) • lim x!2! f(x) • lim x!2+ f(x) • lim x!2 f(x) 8 " .÷ " "" " ÷÷::÷MAC 2311 Learning Activity #1 Intro to Limits Names: Group #: 1. Find the average rate of change of the following functions over the given interval: (a) f(x) = 2x2 ! 3x + 7; [!2, 1] (b) g(t) = 2 cos2(t); h0, ⇡6 i 2. For the graph of f(x) given below, find the following values, if they exist. If it does not exist, state “DNE”. (a) f(!1) (b) lim x!"1 f(x) (c) f(4) (d) lim x!4+ f(x) (e) lim x!4! f(x) (f) lim x!4 f(x) (g) lim x!"4! f(x) (h) lim x!"4 f(x) (i) lim x!"5 f(x) Page 1 of 2 Valentina Espinosa l H - Dftp.f#--8tbt7 -- 21-232-31-2) -17 ARC " "" ¥¥T , = -J ARC -- glbty.gg) 9101=2101401--411 g. ( Mb ) : 2101446) ?g ¥70 - - 21%5=2144) -12 3-4/21 -11ms -31M I -2 3 2 -3 DNE -2 DNE 2MAC 2311 Learning Activity #1 Intro to Limits 3. If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height in meters t seconds later is given by s(t) = 10t ! 2t2. (a) Find the average velocity of the rock over the given time intervals: i. [1, 1.1] ii. [1, 1.01] iii. [1, 1.001] iv. [1, 1.0001] (b) Estimate the instantaneous velocity when t = 1. 4. Complete the table given below for f(x) = x2 ! 1 x ! 1 , then make a conclusion on the given limits. x 0.9 0.99 0.999 1.001 1.01 1.1 f(x) f(x) (a) lim x!1! x2 ! 1 x ! 1 (b) lim x!1+ x2 ! 1 x ! 1 (c) lim x!1 x2 ! 1 x ! 1 (d) Why can’t we just find f(1) to determine lim x!1 f(x)? Page 2 of 2 Varg -_5"= 5.8 Varg -- 54.0155111 = 5.98 Varg -- 511.0011-511 ) TE 5998 Vary -- 511.0001) -511 ) 1¥ - - 5.998 Vins -- hmlsftl - SCU t =6 → ← 1. a 1.991.999/42.001201 lil =L =2 ' " matttfftyam ? =L Stitt't.FI?7a undefinedMAC 2311 Learning Activity #2 Computing Limits Names: Group #: 1. Given lim x!1 f(x) = 3, lim x!1 g(x) = !2, lim x!1 h(x) = 7, compute the following: (a) lim x!1 (f(x) + g(x)) (b) lim x!1 (g(x) ! h(x)) (c) lim x!1 (f(x) · h(x)) (d) lim x!1 (3f(x))2 (e) lim x!1 ✓hg((xx))◆ 2. Consider the piecewise-defined function f(x) = 8<: 0 : x < !5 p25 ! x2 : !5  x < 4 5x : x $ 4 Determine the following limits: (a) lim x!"5+ f(x) (b) lim x!"5! f(x) (c) lim x!"5 f(x) (d) lim x!4+ f(x) (e) lim x!4! f(x) (f) lim x!4 f(x) Page 1 of 3 Valentina Espinosa §M 134-2) 3-2=1 ¥M , ta - G) -9 . ¥7131.17121 ¥M 31312=-92--81 ' x'F , -47 Ex 2=0 5147=20 O 25-1/25=3 Iz --TEST' :O 5147=20MAC 2311 Learning Activity #2 Computing Limits 3. lim x!0 (2x ! 8)1/3 4. lim x!"1 x2 ! 3x ! 4 x2 + 2x ! 3 5. lim x!5 2x2 ! 7x ! 15 x ! 5 6. lim x!1 x ! 1 px ! 1 Page 2 of 3 do -8) "s = -2 th ' - SHH (X-4)¥ Its - - - o #Ht's) ¥712413) 2.5-13 - - IT Hm , lrxtl) = TITI =LMAC 2311 Learning Activity #2 Computing Limits 7. lim x!"1 px2 + 8 ! 3 x + 1 8. lim x!0 e2x ! 1 3ex ! 3 9. lim x! ⇡ 2 1 ! sin2 x cos x Page 3 of 3 ' x'Fitted Mt-is = -Yy ' x'mottled 113641) = 4g final eosin Cork) -- OComputing Limits MAC 2311 Florida International University In this section we will discuss techniques for computing limits for di↵erent types of functions and present a few basic rules to help us simplify the computation of limits. 1 Limit Laws Assuming the limits lim x!a f(x) and lim x!a g(x) both exist, then: • Sum: lim x!a (f(x) + g(x)) = lim x!a f(x) + lim x!a g(x) • Di↵erence: lim x!a (f(x) ! g(x)) = lim x!a f(x) ! lim x!a g(x) • Constant Multiple: lim x!a cf(x) = c lim x!a f(x), where c is a constant • Product: lim x!a (f(x) · g(x)) = lim x!a f(x) · lim x!a g(x) • Quotient: lim x!a f(x) g(x) = lim x!a f(x) lim x!a g(x) (provided the limit of lim x!a g(x) 6= 0) • Power: lim x!a (f(x)) nm = ⇣xlim !a f(x)⌘ nm , where n, m are integers with no common factors and m 6= 0 Note: These laws are also true for one-sided limits as x ! a" and x ! a+. Example 1: Given lim x!2 f(x) = 5, lim x!2 g(x) = 8, lim x!2 h(x) = 2, compute lim x!2 f(x)2 g(x) ! 3h(x). In your work, be sure to indicate each use of a limit law. lim x!2 f(x)2 g(x) ! 3h(x) = 1 limits2 Common Limits • The limit for a constant function, f(x) = c, for any constant c: lim x!a f(x) = lim x!a c = c regardless of the value of a • The limit for the identity function, f(x) = x: lim x!a f(x) = lim x!a x = a • The limit of a polynomial function, p(x) = cnxn + · · · + c1x + c0: lim x!a p(x) = cnan + · · · + c1a + c0 = p(a) • The limit of a rational function (the quotient of two polynomials): lim x!a p(x) q(x) = p(a) q(a) if q(a) 6= 0. 3 Techniques for Computing Limits 3.1 Direct Substitution When presented with any limit, direct substitution is always the first step in trying to evaluate the limit. Direct substitution is simply plugging the x-value being approached directly into the function. Example 2: • lim x!2 (x3 ! x) = 23 ! 2 = 8 ! 2 = 6 • lim x!0 ex = e0 = 1 • lim x! ⇡ 2 cos(2x) • lim x!1 x x + 2 • lim x!3 100 2 plug -- cos 12.42) : - I ii.mi it 1003.2 Evaluating Limits of Functions as Quotients Just as any other type of limit, when evaluating a limit that contains a quotient, trying direct substitution is the first step. If you substitute in the x-value that is being approached and the result is a real number, then you are done. However, in the event that the result of direct substitution yields an invalid output, there are some algebraic techniques we can use to determine the value of the limit, if it exists. For rational functions, if after trying direct substitution you get a result of 0 0 (an indeterminate form), then try factoring the numerator and/or denominator or try simplifying in order to find a common factor that can be canceled before trying direct substitution again. Example: Evaluate the following: lim x!1 x2 + x ! 2 x2 ! x Let’s try direct substitution first: 12 + 1 ! 2 12 ! 1 ! 0 0 Since direct substitution results in the indeterminate form of 0 0 , let’s try factoring the expression and canceling before utilizing direct substitution again. lim x!1 x2 + x ! 2 x2 ! x = lim x!1 (x !x(1)( x !x1) + 2) = lim x!1 x + 2 x = 1 + 2 1 = 3 Example: Evaluate the following: lim x!3 1x ! 13 x ! 3 Direct substitution: 13 ! 13 3 ! 3 ! 0 0 Simplify/factor expression: lim x!3 1x ! 13 x ! 3 = lim x!3 3 3x ! x 3x x ! 3 = lim x!3 3"x 3x x ! 3 = lim x!3 3 ! x 3x(x ! 3) = lim x!3 !1(x ! 3) 3x(x ! 3) = lim x!3 !1 3x = !1 3 · 3 = ! 1 9 3Example 3: Evaluate the following: lim x!"3 x2 + 4x + 3 2x + 6 Example 4: Evaluate the following: lim x!2 x3 ! 8 x2 ! 4 Example 5: Evaluate the following: lim x!4 1 ! 4 x 4 ! x 4 Simplify Htt ) ftp.sltlxth) (H - Stl)) = - I simplify : X42Xt4 Xt 2 MY mitts (" ) -- 24¥24 is simplify LIII -- Ix plugin : -114For limits of quotient functions involving a square root in the numerator or denominator, if direct substitution leads to the indeterminate form of 0 0 , then we should try using the conjugate to rationalize the expression in order to find a common factor to cancel before utilizing direct substitution again. Example: Evaluate the following: lim x!4 x ! 4 px ! 2 Direct substitution: p44!!42 ! 0 0 Let’s try using the conjugate to rationalize the expression, cancel a common factor, and then direct substitution. lim x!4 x ! 4 px ! 2 = lim x!4 pxx!!42 · p px x + 2 + 2 = lim x!4 (x ! 4)(px + 2) x ! 4 = lim x!4 (px + 2) = p4 + 2 = 4 Example 6: Evaluate the following: lim x!1 px + 8 ! 3 x ! 1 5 rationalre :µ¥ pwg in im , Ixr.tt) = 46 [Show More]

Last updated: 2 years ago

Preview 1 out of 45 pages