Hw2 53 Answer Key

ANSWER KEY ( HOMEWORK # 2 ) 1. (a) KN

(b) D.N.E.

2 x2 2. (a) f x = 2 x K1

(c) 1 lim

x/1 C

lim

(d) 1

(e) 1

(f) 0.5

f x =CN, lim f x =KN x / 1K

x /K1 C

f x =KN, lim

f x =CN

x /K1K

(g) 0

(h) D.N.E.

Hence x = 1 is a VA Hence x =K1 is a VA

x C1 f x is continuous on its domain which is KN,K1 W K1, 1 W 1,CN x K1 sin π$x , x 2 0, 1 (c) f x = x$ x K1 sin π$x sin π$x π Notice that limC = limC $ =Kπ x/0 x$ x K1 x/0 x K1 π$x Also sin π Kπ$x = sin π$x , so we have : sin π$x sin π$ 1Kx Kπ sin π$ 1Kx lim = lim = lim $ =Kπ K K K x$ x K1 x$ x K1 x x/1 x/1 x/1 π$ 1 Kx Hence define f 0 = f 1 =Kπ so that f x is continuous on 0, 1 (b) f x =

Kx2

3. Let f x = e

Kx. Clearly f x is continuous on =. Now, f 0 = 1 O 0 and f 1 =

means that d a constant c 2 0, 1

2

such that f c = 0 by the IVT. Since f c = 0, 0 eKc = c

4. (a) TRUE. Given f x , horizontal asymptotes are determined by computing lim

x /CN

(b) FALSE. Take f x =

x C1, x s 0 2,

1 K1 ! 0 e

x=0

f x or lim f x x /KN

For all x, f x O 1 but lim f x = 1

π π (c) FALSE. Take f x = tan x , x 2 K , 2 2

x/0

f x is continuous on this interval but we won't be

π π able to find a function g x continuous on = which satisfies g x = f x for all x 2 K , 2 2 1 sin x 1 sin x cos x cos x

sin x 2 x Ctan x

5. (a) lim

x/0

(b) lim

1Ktan x cos x Ksin x

(c) x /CN lim

3x K3Kx 3x C3Kx

π x/ 4

(d) lim

x /KN

2 xC

3Kx 3Kx

= lim

x/0

= lim

π x/ 4

= x /CN lim

4 x2 C3 x

1 x 1 2$ C sin x cos x cos x Ksin x cos x Ksin x $cos x

=

=

1 3 2

1 K3K2 x =1 1 C3K2 x 2

2 xK 4 x C3 x 2 xK 4 x2 C3 x

K3 x

= lim

x /KN

2 xK 4 x2 C3 x

1 x 1

K

x2 = lim

x /KN

3 K2K 4 C

3 4

=K 3 x

6. Prove that if xlim f x = L, then xlim f x /a /a

= L

Proof : Let g x = x which is continuous on =. Then g f x = f x Now, xlim f x = xlim g f x = g xlim f x =g L = L /a /a /a 7.

f x =

0,

x is rational

x,

x is irrational

Is f x continuous at x = 0 ? at x = 1 ?

f x is continuous at only one point, at x = 0 Proof : For x = 0, given any e O 0, take δ = e. Then if 0 ! x K0 ! δ = e, we have for any rational x in the δKneighborhood that f x = 0 ! e , while for any irrational x, f x = x ! e. This shows that the lim f x = 0 = f 0 . Hence, f x is continuous at x = 0. x/0

Now, for any other number a s 0, xlim f x does not exist. For the sake of argument, suppose the /a f x = L. Now we can't have L = 0 since by taking e = a O 0, we have limit does exist and let xlim /a that for any δKneighborhood about a, then 0 ! x Ka ! δ 0 d x irrational in the neighborhood such that f x K0 = x O a = e. Similarly, if L s 0, by taking e = min L , a , then for any δKneighborhood about a, then 0 ! x Ka ! δ 0 d x rational in the neighborhood such that f x KL = 0 KL = L R e. So indeed xlim f x D.N.E. and f x is not continuous for a s 0. /a