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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. Mark 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, x2 is . . . . . . . 2 . x2—3 x+2 _
negative, x1 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 (2111) —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(2ih] +\/ 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 poinTslope formula) y— 2 = %(x— 2) or y=§+1. x Plot[{m, —+1},{x, o, 4}] 2 3.07 ...
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 Spring '14
 DragaVidakovic
 Geometry

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