200701 Umaths Tma2 Ansguide

WUC 112 University Mathematics (January 2007) TMA2 – Answer Guide Question 1 x 2 + 2x − 3 (a) lim x →1 x −1 ( x + 3)( x − 1) = lim x →1 x −1 ( x + 3) = lim x →1 =1+3=4 1 1 1 −   (b) lim x →0 x 2 + x 2  1 −x    = lim x →0 x  2( 2 + x )    −1 = lim x →0 2( 2 + x ) 1 =− 4 x−4 (c) lim x→ 4 x −2

 x  2  x  2  x  2 ( x  4)  x  2  = lim = lim x4

( x  4)

( x  4)

x4

 lim x  2  4 x4

x  1, x  1  0, x  1  ( x  1), x  1  0, x  1 x 1 x 1 , lim−  lim+ x →1 x →1 x 1 x 1 x −1 therefore lim does not exist. x→1 x − 1 

(d) x  1  

Question 2 3 f ( x) (a) lim x→c g ( x ) f ( x) = 3 lim x→c g ( x )  7  =3    −3  = −7

9 − 2 x 2 ≤ g ( x) ≤ 9 − x 2 ,

(b) Given

lim 9 − 2 x 2 = 3 and lim 9 − x 2 = 3 x →0 x →0 g ( x) = 3 Thus, lim x →0 1 1 = g ( x) 3 2 x x2 (c) Given 1 ≤ f ( x) ≤ 1 + 4 2 2  x2 x 1 + lim (1 − ) = 1 and lim x →0  x →0 2 4  Thus, lim f ( x) = 1 and lim x →0

  = 1 

x→0

2 f ( x) + 1] = 2(1) + 1 = 3 and lim[ x →0 f ( x) = 0 (d) Given lim x→c Since - f ( x) ≤ f ( x ) ≤

f ( x)

lim − f ( x) = lim f ( x ) = 0 x →c x →c f ( x) = 0 Therefore, lim x →c Question 3

2 x −1 + 2 x x −1 2( x 2 − 1) + x( x − 1) = x( x 2 − 1) x( x 2 − 1) ≠ 0 , thus, x ≠ 0 and x 2 ≠ 1 f (x) is not continuous at x = 0, −1 and 1.

(a) f (x ) =

(b) f (x) =

x 4x + x 2

.

4 x 2 + x ≠ 0 , thus, x ≠ 0 and 4 x + 1 ≠ 0 1 f (x) is not continuous at x = − and 0 4 x−2 (c) f (x) = . x −2 x − 2 ≠ 0 , x ≠ 2 thus, x ≠ 2 f (x) is not continuous at x = ± 2

(d) f (x) =

x+8 x 2 + 8x

,

x( x + 8) ≠ 0 thus, ( x + 8) ≠ 0 and x ≠ 0 f (x) is not continuous at x = −8 and 0 Question 4 (a) Let f(x) = x3 + x2 – 2x – 1 f (–1) = (–1) 3 + (–1) 2 – 2 (–1) –1 = 1 > 0 and f (1) = (1)3 + (1)2 – 2(1) – 1 = –1 < 0 Thus, there is at least one solution in [ -1, 1 ] (b) Given f (x) = f(2)=

1 1 + x −1 x − 4

1 1 > 0 and f ( 3 ) = − < 0 , 2 2

therefore, there is a number c ∈ ( 1, 4 ) such that f (x) = 0. (c) Let F(x) = f(x) – 1 = x5 – 2x2 + 5x – 1, F (0) = –1, F (1) = 3. Thus, there is a number c in ( 0, 1 ) such that F (x) = 0 which implies that f (c) = 1 Question 5  x, x ≥ 0 (a) f ( x ) = x =  − x, x < 0 lim− f ( x) = lim+ f ( x) = f (0) = 0 x →0

x →0

Thus x is continuous for all x. x ≤1 3 x − 2, (b) (i) f ( x ) =  2 x >1  kx ,

f ( x) = lim+ f ( x) The function is continuous at x = 1, if and only if xlim →1− x →1 lim f ( x) = 3(1) − 2 = 1 ,

x →1−

lim f ( x) = k (1) 2 = k ,

x →1+

Thus, k = 1.  kx 2 , g ( x ) = (b) (ii)  2 x + k ,

x≤2 x>2

g ( x) = lim+ g ( x ) The function is continuous at x = 2, if and only if xlim → 2− x →2 lim g ( x) = 4k ,

x →2 −

lim g ( x) = 4 + k ,

x →2 +

4k = 4 + k, thus k =

4 3

Question 6 tan x sin x = lim x →0 x →0 x cos x x  sin x  1  = lim    x →0  x  cos x   sin x   1  = lim  lim    x →0  x  x →0  cos x  = (1) (1) =1

(a) lim

sin 2 x 2sin x cos x = lim x →0 x →0 3x 3x 2 sin x    lim  lim cos x 3  x 0 x  x 0 2 = 3

(b) lim





1  cos x 1  1  sin 2 x = lim x 0 x 0 x x

(c) lim

= lim x 0 = lim x 0 = lim x 0

1  1  sin 2 x  1  1  sin 2 x     1  1  sin 2 x  x   1  (1  sin 2 x )



x 1  1  sin 2 x



sin 2 x

x 1  1  sin 2 x

 

sin x sin x lim = x 0 x x 0 1  1  sin 2 x lim





= (1)(0) = 0 tan 2 x tan 2 x = lim x 0 sin 3 x x 0 sin 3 x sin 2 x = lim x 0 (cos 2 x ) sin 3 x 2 sin x cos x = lim x 0 (cos 2 x )(2sin x cos 2 x  cos 2 x sin x )

(d) lim

= lim x 0 =

2 cos x (cos 2 x)(2 cos 2 x  cos 2 x) 2 lim cos x x 0

(lim cos 2 x)(2 lim cos 2 x  lim cos 2 x ) x 0

x 0

x 0

2(1) 2  = (1)(2(1)  (1)) 3 Question 7 (a) Let ε >0 and there exist δ >0 such that 0< x − 4 < δ , ⇒

x 2 − 16 −8 < ε . x−4

x 2 − 16 −8 < ε x−4 ⇒

x 2 − 16 − 8( x − 4) <ε x−4

⇒

x 2 − 16 − 8 x + 32 <ε x−4

⇒

x 2 − 8 x + 16 <ε x−4

( x − 4)( x − 4) <ε x−4 ⇒ x−4 < ε Hence, we can let δ = ε ⇒

(b) Let ε > 0 , we seek a number δ > 0 such that if 0 < x − 2 < δ , then, (2 x − 3) − 1 < ε . (2 x − 3) − 1 < ε ⇒ 2x − 4 < ε ⇒2 x −2 < ε , 1 ⇒ x−2 < ε 2 1 thus we have δ = ε 2 (c)

 x − 5, x ≥ 5 x −5 =  5 − x, x < 5 lim f ( x) =

x →5−

( x + 2)( x − 5) −( x + 2)(5 − x) = = −( x + 2) = −7 , (5 − x) (5 − x)

( x + 2)( x − 5) = ( x + 2) = 7 x→5 ( x − 5) lim f ( x ) ≠ lim+ f ( x ) , therefore lim f ( x) does not exist. x→ 5 x →5 − x →5 lim+ f ( x) =

Question 8  2 f ( x)   −2    = 2   = -1 (a) lim x →c  3 + 1  g ( x) + 1  (b) If f (x) is continuous at x = - 3 , -1 = p + q(-3) that is p – 3q = -1 ……………(1) If f (x) is continuous at x = 0, p + q(0) = 5 (0) that is p = 0 …………………(2) Solving, (1) and (2) 1 we obtain p = 0 and q = 3

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