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F08 Math 1205 -- Test 1 17 Sep 2008 NAME: SOLUTIONS
Justify all work using complete sentences! Use only methods from class.
x
fHxL
- [12] a) Let f H xL be defined by the graph given above. Determine the following limits, writing +¶ or - ¶ as appropriate. If a limit does not exist, explain why.
lim
xØ-¶
f HxL = 0 lim
xØ- 2
f HxL = 1 lim
xØ 4
f HxL = 1
lim
xØ- 3 -
f HxL = - 3 lim
xØ- 3 +
f HxL = 1 lim
xØ- 3
f HxL DNE
lim
xØ- 3 -
f HxL ≠ lim
xØ- 3 +
f HxL
lim
xØ 1 -
f HxL = +¶ lim
xØ 1 +
f HxL = - ¶ lim
xØ 1
f HxL DNE
lim
xØ 1 -
f HxL ≠ lim
xØ 1 +
f HxL
lim
xØ 3
f HxL = +¶ lim
xØ 2
f HxL = 0 lim
xØ- 1
f HxL DNE
lim
xØ- 1 -
f HxL ≠ lim
xØ- 1 +
f HxL
- [4] b) Using the graph of f above, label the following as True/False. i) F_ f H xL is discontinuous at x = - 3 since f H xL is not defined at x = - 3. ii) F_ f H xL has exactly 2 removable discontinuities. iii) T_ y = 0 is a horizontal asymptote of f H xL. iv) F_ f H xL has a jump discontinuity at x = 4.
- [20] Evaluate the following limits using +¶, - ¶ and “does not exist”, where applicable. Use only algebraic methods. Tables alone will not suffice.
a) lim tØ 2 -
3 t^2 - 6 t … t- 2 … =^ lim tØ 2 -
3 tI t- 2 M
- I t- 2 M Hsince^ t^ -^2 <^0 when^ t^ Ø^2
- L
= lim tØ 2 -
= - 6
b) lim xØ 0
3 tanI 2 xM sinH xL =lim xØ 0
3 sinI 2 xM cosI 2 xM sinH xL KtanH^2 xL^ =^
sinI 2 xM cosI 2 xM O = lim xØ 0
3 cosI 2 xM
I 2 xM sinI 2 xM 2 x
x x sinH xL Jmultiply^ by^
2 x 2 x and
x x to^ get^ in^ correct^ formN = lim xØ 0
3 I 2 xM x
sinI 2 xM 2 x
x sinH xL Hreorganizing^ termsL
= 6 * 1 * 1 since lim xØq
sinHqL q =^1 = 6
c) lim hØ 0
h^2 h^2 + 12 - 12
= lim hØ 0
I h^2 M h^2 + 12 + 12 h^2 + 12 - 12 h^2 + 12 + 12
Hmultiply by conjugateL
= lim hØ 0
I h^2 M h^2 + 12 + 12 h^2 + 12 - 12 Hexpand^ denominatorL
= lim hØ 0
I h^2 M h^2 + 12 + 12 h^2 HsimplifyL
= lim hØ 0
h^2 + 12 + 12 Icancel h^2 M
= 2 12 = 4 3 d) lim xØ 3 k x + 2 ( k is a constantL = 3 k + 2
- Given: f H xL = x^ x 2 - -^24
a) [6] Determine all vertical and horizontal asymptotes of f H xL. H.A.: lim xر¶
x- 2 x^2 - 4 =^ xlimر¶
x x^2 -^ 2 x^2 x^2 x^2 -^
4 x^2
= lim xر¶
x^1 -^2 x^2 1 - (^) x^42 = 01 - - 00 = 0 ï y = 0 is a H. A.
V.A.: f H xL = x^ x 2 - -^24 = I x- 2 xM- I^2 x+ 2 M lim xØ- 2 -
x- 2 I x+ 2 M I x- 2 M =^ xlimØ- 2 -
1 x+ 2 =^ - ¶^ H^ x^ +^2 <^0 when^ x^ Ø^ -^2
- L
lim xØ- 2 +
x- 2 I x+ 2 M I x- 2 M =^ xlimØ- 2 +
1 x+ 2 =^ +¶^ I^ x^ +^2 >^0 when^ x^ Ø^ -^2
+M
So, x = - 2 is a V. A. NOTE: x = 2 is NOT a V. A. since lim xØ 2
f H xL ≠ ±¶ b) [4] For each vertical asymptote, determine whether f H xL Ø +¶ or f H xL Ø - ¶ on either side of the asymptote.
x = - 2 is the only vertical asymptote lim xØ- 2 -
x- 2 I x+ 2 M I x- 2 M =^ xlimØ- 2 -
1 x+ 2 =^ - ¶^ H^ x^ +^2 <^0 when^ x^ Ø^ -^2
- L
lim xØ- 2 +
x- 2 I x+ 2 M I x- 2 M =^ xlimØ- 2 +
1 x+ 2 =^ +¶^ I^ x^ +^2 >^0 when^ x^ Ø^ -^2
+M
