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Unformatted text preview: Math 1205  Test 1 20 Sep 2006
f Hx L
5 4 3 2 1 NAME: Explain all work! Use only methods developed in class. 6 5 4 3 2 1 1 2 3 4 1 2 3 4 5 6 x 1. [16] a) Let f Hx L be defined by the graph given above. Determine the following limits, writing + or  as appropriate. If a limit does not exist, explain why. x  lim f Hx L lim x ض x H1L x 1 lim f Hx L f Hx L 4 4 HDNEL 2 lim f Hx L lim x H1L+ x 1+ x 2  lim f Hx L lim f Hx L f Hx L 4 2 HDNEL 2 x 3 x 1 x 1 x 2 + lim f Hx L lim f Hx L DNE left right HDNEL 2 lim f Hx L lim f Hx L 2 x 2 lim f Hx L (1 continued) b) Using the definition of asymptotes, explain why x = 1 is a vertical asymptote, but x = 1 is not. Vert. Asymps( lim
x a f Hx L = ): x = 1 satisfies the definition; f Hx L is finite near x = 1 (1 continued) b) Using the definition of asymptotes, explain why x = 1 is a vertical asymptote, but x = 1 is not. Vert. Asymps( lim
x a c) Give the x values for which f is not continuous. Explain the kind of discontinuity of each. At x = 1 left and right limits exist but are unequal(jump discontinuity); at x = 1 the limit does not ity). f is continuous at x = 3 exist; at x = 2 left and right limits exist and are equal but are not equal to f H2L (removable discontinu2. [8] Can f Hx L approach a limit as x c even if f Hc L is undefined? If so, give an example.
sinHx L Yes, f Hx L = with c=0 is an example. lim f Hx L = 1, but f is undefined at x = 0. f Hx L = ): x = 1 satisfies the definition; f Hx L is finite near x = 1 x x 0 logHx +1L x 3. [8] If we know that 1  1 + x for x .5, then show how to use the Squeeze Theo rem to find lim . lim f Hx L = b .
x a The Squeeze Theorem tells us that if g1 Hx L f Hx L g2 Hx L near a and lim g1 Hx L = b = lim g2 Hx L, then
x a x a logHx +1L x x 0 2 x Let x lim g1 Hx L = lim I1  M = 1 = lim H1 + x L = lim g2 Hx L, then lim f Hx L = 1.
2 x 0 x g1 Hx L = 1  x 0 and 2 x 0 g2 Hx L = 1 + x ,
x1 x>1 a=0, we see that since x 0 x 0 4. [8] Let f Hx L 2x 2 x +2ax +2 . What should a be to make the function continuous?
x 1+ f Hx L will be continuous at x = 1 if lim f Hx L = lim Ix 2 + 2 a x + 2M = 3 + 2 a and lim f Hx L = lim 2 x = 2,
exist and are equal. We need 3 + 2 a = 2, or a
x 1+ 1 =  . 2 x 1x 1 Determine lim f Hx L and lim f Hx L.
x 1+ x 1 5. [8] If lim f H g Hx LL exists, must lim g Hx L also exist? Explain or give a counterexample.
x 0 x 0 x 1 1 1 all what g Hx L is! Consider also f Hx L = g Hx L = ; f H g Hx LL = = = x , which gives another counterex1 ample. g 's limit need not exist. Suppose, for example, f Hx L = 1; the first limit exists and it doesn't matter at
g Hx L x 6. [28] Evaluate the following limits using +,  and "does not exist" where applicable. Use only algebraic methods (i.e., no tables). Justify your answers.
H2 x +3L 2 x2 2 x +3 7 a) lim x 6 = lim Hx 2L = lim = 2 x 2 3 x 7 x +2 b) Identify the horizontal and vertical asymptotes of the function in a)
1 Using the factorization and result of a), we see that x = 2 is not an asymptote, but x = is since the 55 1 2 x 2 x 6 2 2 3 x 2 H3 x 1L Hx 2L x 2 3 x 1 5 6. [28] Evaluate the following limits using +,  and "does not exist" where applicable. Use only algebraic methods (i.e., no tables). Justify your answers.
H2 x +3L 2 x2 2 x +3 7 a) lim x 6 = lim Hx 2L = lim = 2 x 2 3 x 7 x +2 b) Identify the horizontal and vertical asymptotes of the function in a)
1 Using the factorization and result of a), we see that x = 2 is not an asymptote, but x = is since the x 2 H3 x 1L Hx 2L
1 . 3 x 2 3 x 1 5 numerator is c) lim
2 x 2  d) x 1 x 1 x +1 lim = x 1+ x 1 h0 h x +1 lim Hh+1L3 1 h3 +3 h2 +3 h+1  1 h3 +3 h2 e) lim = lim = lim +3 h = lim Ih2 + 3 h + 3M = 3 (because x  2 = Hx  2L when x < 2) 3 x +4H 2L = lim  x = lim  3 x + 4 = 10. x 2 x 2 x 2 Check right and left limits:
h0 = 3 x 2 x 8 x 2 55 9 0 2 x2 2 2 at x = Since lim x 6 = , y = is the only 3 x 2 7 x +2 3 3 x ر H3 x +4L H x 2L 3 x +4H 2L 3 x +4 x 2 lim = lim = lim  x x 2 x 2 x 2 x 2 x 2 x 2 horizontal asymptote. 3 x +1 lim x 1 x 1 =  since the denominator is negative and since the denominator is positive. The limit does not exist. f) g) Let f Hx L =
x a of a and the limit. x 2 lim !!!!!!!!!! x 1 1 x 2 x +a x lim f Hx L = lim = lim +a . We can see that the limit of the denominator will be 2 zero () and therefore the whole limit will not exist unless possibly the limit of the numerator is also zero. The limit of the numerator:
x a
2 x a x  Ha  1L x  a h h h0 h0 !!!!!!!!!! !!!!!!!!!! !!!!!!!!!!! Hx 2L x 1 +1 x 2 x 1 +1 = lim = lim !!!!!!!!!! = lim x  1 + 1 = 2 !!!!!!!!!! Hx 1L1 x 1 1 x 1 +1 x 2 x 2 x 2 x 2 +a . Are there any values of a such that lim f Hx L exists? 2 x  Ha  1L x  a x a
2 If so, give the value x a Hx  a L Hx + 1L 2 x x If a = 0 lim f Hx L = lim = lim = 0 2 If a = 1 lim f Hx L =
x 0 x 1 lim Ix 2 + a M = a 2 + a = a Ha + 1L is only zero if a = 0 or 1.
x 0 x +x x 0 x +1 Hx 1L Hx +1L x 2 1 lim = lim 2 2 Hx +1L x 1 x +2 x +1 x 1 x 1 = lim Does not exist (see d)). x 1 x +1 Summing up, if a = 0, the limit is 0, otherwise the limit does not exist. 7. [7] Show that f Hx L 2 x 3 + x 2 + 2 has a root in H 2, 1L. Name any theorem(s) used and show all steps necessary. (Do not attempt to find a decimal approximation to the root.) We use the Intermediate Value Theorem. The function is continuous; f H2L = 10 and f H1L = 1. Zero is an intermediate value between 10 and 1, so by the theorem there exists an x* oe H2, 1L such that f Hx* L = 2 x* 3 + x* 2 + 2 = 0 as requested. choice. Since: 5 x  10 < 0.2 if and only if 5 x  2 < 0.2 if and only if x  2 < .04 8. [8] What is the largest value of d that guarantees " x  2 < d implies 5 x  10 < 0.2 "? Justify your Hence d = .04 guarantees that x  2 < d implies 5 x  10 < 0.2 and no smaller d will do. 8. [8] What is the largest value of d that guarantees " x  2 < d implies 5 x  10 < 0.2 "? Justify your choice. Since: 5 x  10 < 0.2 if and only if 5 x  2 < 0.2 if and only if x  2 < .04 9. [9] Let f Hx L The
f H4+hL H4L lim f 4+h4 h0 Hence d = .04 guarantees that x  2 < d implies 5 x  10 < 0.2 and no smaller d will do. a) Find the slope of the secant line through the points H4, f H4LL and H5, f H5LL. !!!! !!!! f H5Lf 5 2 The slope of the secant is H4L = = 5  2 54 1 b) Find the slope of the tangent line at the point H4, f H4LL by evaluating a limit (show work). slope of c) Find the equation of the tangent line through H4, f H4LL.
4 1 1 Using the pointslope formula at (4,2): y  2 = Hx  4L fl y = x + 1 4 !!!!!!!!!! 4+h = lim 2 h h0 !!!! x. the tangent = !!!!!!!!!! 4+h lim 2 h h0 line at !!!!!!!!!! 4+h +2 !!!!!!!!!! 4+h +2 (4,f(4)) H4+hL4 = lim !!!!!!!!!! h0 h 4+h +2 is by 1 1 = lim = !!!!!!!!!! definition
h0 4+h +2 4 ...
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 Fall '08
 FBHinkelmann
 Math, Calculus

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