Review problems - You will be graded on how well and fully...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: You will be graded on how well and fully you support your work! Use only methods from class. F05) l 2 3 4 5 6 1. [24] a) Let f(x) be defined by the graph given above. Mar-k each statement true or' false. f(x)is continuous lirn f(x) exists lirn f(x) = l 01. x = 1 x—>3 x—>2‘ 2 TRUE and is = +00 FALSE TRUE W :1 lim f(x) exists x: 3isaver'tical Iim f(x) at Iim f(x) 3 3H5 asymptote “'3— X->3+ TRUE TRUE TRUE TRUE f (x) is discontinuous x = 5 is a removable Iim f (x) does not exist x = 4 is a removable at x = 1, 2, 3, 4, 5 discontinuity off x43 discontinuityoff TRUE TRUE FALSE FALSE 2 2. [9] Determine the value(s) of a for‘ which Iim X fax?” must exist. Evaluate the limit in this case(s). X62 X— x2—5 x+6 = (x72)(x73) x = 2 as well. Thus, 22 — 2 a + a +1: 0, or after rearranging, a=5. The expression becomes: 2 2 , so the X— X— 2 2 _ _ _ limit exists for this value. Iim L"+‘”1=lim fl = lim W = Iim (x— 3): —1 va X_2 xaz X_2 xaz X_2 va 3. [9] Suppose that (x — 1)2 5 fix) 5 cos % for all x e [0, 1]. For what value(s) of c must lim fix) exist? Determine the XHC corresponding limit value(s). It is apparent from the setup that the Sandwich Theorem is called for. To see where the pinching occurs, let's do a quick plot: The blue curve is cos(7r x/2) and the red curve is (x—1)2. . | . l . | . l . l l l l 0.2 0.4 0.6 0.8 1.0 It's clear from the graph that sandwiching occurs only at x=O and x=1. In fact, the hypothesis of the Sandwich Theorem apply at these two points. Finally, limX_)O fix) = 1 and limX_,1 fix): 0 4. [21] Evaluate the following limits. Use only algebraic methods (i.e., no tables, no decimals). A bare answer is not adequate; Justify your work. 1; 3 2 a) lim Xsixz = lim ’1 = l by the basic limit laws theorem and the fact that lim 1 = 0. xaioo 2 X +X Xaioo 2+; 2 x4790 X . X273 Hz . (x72) (x71) . . . . . . . . b) lIm — = lIm — Since the denomlnator vanishes at the lImIt pomt Wl‘l'l'IOLl‘l' the benefit of the numera- X_)_2+ x2(x+2) x2(x+2) x—>—2+ tor vanishing, the limit does not exist. We need only check whether its -00 or +00. At x a little greater than -2, x-2 is . . . . . . . 2 . x2—3 x+2 _ negative, x-1 Is negative and x+2 Is posmve as Is x . Thus llm 2—-+oo. X_)_2+ x (x+2) c) lIm [Hint: cos(2 x) = 1— 2 sin2(x)] Using the hint (sorry about that typo), we have: x—>O I xsin(7rx)_|irnl xsin(7rx) xsin(7rx) _|im1 x simrx 7r x simrx x lim _ 2 =lim _ _ — _ _ = lim — _ _ As x—>O, each of the three factors X60 licos(2 x) X40 25m (x) X_)0 ZSInxsmx X60 2 sinx sinx X60 2 sinx 7rx smx on the right tends to 1! (As derived in class.) Therefore, Iim % = XHO 5. [9] Show that there is an x e [0, which satisfies sin(x) = (X. [Hint: Use the Intermediate Value Theorem.] To use the IVT, we must identify a function to apply it to. Let, fix) = sinx — 13’" so that we are seeking a solution to fix) = 0. Evaluating at the endpoints of the interval, we see fiO) = sinO — e0 = 0—1 = —1 and = sin? — e_§ =1— L, > 0, this 55 last since we know e5 >1. Thus, fix) is negative at x=0 and positive at x=1, which means by the IVT that there is an xoin (o, with 71m) = o. 6, [9] Suppose we know that, for any 6 > 0,0 < | x— )@| < 6 implies that l fix) — f(x0)| < Is fix) continuous at )q)? fix) is con1'inuous a1' x0 if lim fix) = fixo). For 1'his limi1' 10 hold we need 10 show 1'ha1' for any e > 0, 1'here is a 6 such x—mo 2 1'ha1' O<|x—)q)| <6irnplies |f (x) —f (x0) | < e. Le1' e>0 be given and se1'6=e. 5uppose0<|x—x0| <6: by hypo1hesis 1'his implies | fix) — flx0)| < V— = V 62 = e, which is exac1'ly wha1' we needed 10 show. 2 7. [10] Find all asymp1'01'es and poin1's of discon1'inui1'y of fix) = x + x7 5 4 17—+— - - x2—5 x+4 - X x2 - - Horizon1'al asymp1o1'es: llm — = llm 3 4 = 1 —> y: 1 Is 1'he only horizon1'al asymp1'o1'e X—>+/—oo X2+3X’4 X—>+/—oo 1+;7— Discontinui1'ies: The only discon1'inui1'ies for a ra1'ional func1'ion are 1'he zeroes of 1'he denomina1'or: den = (x+ 4) (x — 1) —> x = 1 and x = —4 are poin1's of discon1'inui1'y. Ver1'ical asymp1'o1'es: As wi1h discon1'inui1'ies we search for 1he zeroes of 1'he denomina1'or: x = 1, —4 We mus1' also now check whe1'her 1'he numera1'or also vanishes of one of 1'hese poin1's: num = (x —4)(x— 1). Thus, a1 x=1 1'he zero in 1'he denomina1'or is cancelled by 1'he zero in 1'he numera1'or. The only ver1'ical asymp1'01e is x=-4. 8. [9] Le1' fix) = V 2 x . De1'errnine f' (2) direc1'ly using 1'he difference quofiem‘ definifion Find The equa1'ion of 1'he 1'anen1' line 10 fix) 01' x = 2. Sketch 1'he curve and 1'he 1anen1' line. ‘1 2 (2-1-11) —1j 2 x2 1'? f'(2)=lim mam. = ll m h—»O 1’ m0 h—>O \l 2 (2+h) —\,‘ 2 X2 x ‘l 2(2-i-h] +\/ 2x2 h = “m 2[2+h)—4 hso h(‘/2(2+h) +\/2><2) =lim 4: haO 2(2+h) +2 The equa1ion of The line through 1'he poin’r, (2, 2), with slope m: % is (using poinT-slope formula) y— 2 = %(x— 2) or y=§+1. x Plot[{m, —+1},{x, o, 4}] 2 3.07 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern