d. f(0) =

= limx-+0

= limx-t 0

rim !d:{Qh+0 x-u

x-sin2xsln x (-1)

xx-sin2x+sinx

xstnxUsing TABLE for numerator, denominator, andquotient shows that the numerator goes to zerofaster than the denominator. For instance, ifx = 0.00'l ,

. 1.1666... x 10-9 _ ^ ^^r {aQuotient = = 0.001 16... .

9.999... x 10-7Thus, the limit appears to be zero.(The limit can be found algebraically to equal zero byI'Hospital's rule after students have studied Section6-8.)

Chopter TestT1. See definition of limit in Chapter 2.T2. See definition of derivative in Section 3-2 or 3'4.T3. Prove that if f(x) = 3t', then f'(x) = llx3.

Proof :

*;,l.a"tire=,1'lrg#= lim

3x4 + 12x3h + 18x2h2 + 12xh3 + 3h4 - 3x4

h-+0

= lim (12x3+

T . t(x) = cos 3x + t'(x) = -3 sin 3x.t'(5) = -3 sin 15 = -1.95086...Decreasing at 1.95... y-units per x-unit.

T5. Graph (below). lf you zoom in on the point where x = 5,the graph appears to get closer and closer to thetangent line. The name of this property is locallinearity.

T6. Amos substituted before differentiating instead olafter.Correct solution is f(x) = 7x + f'(x) =7 + f'(5) = 7.

h

18x2h + 12xh2 + 3h3; = 12xs, Q.E.D.

T7. Graph. f(x) = i!!I

As x approaches 0, f(x) approaches 1.The squeeze theorem. lt states that:

lf (1) g(x) < h(x) for all x in aneighborhood of c,

(2) ,lg s(x) = Jg

h(x) = l-, 6n6

(3) f is a function for whichg(x) < f(x) < h(x) for all x in thatneighborhood of c,

Then lim f(x) = L.

T8. f(x) = (7x + 3)1s - f'(x) = 105(7x + 3)14

T9. g(x) = cos (x5) + g'(x) = *5x4 sln xs

tto. *q{sin 5x) = 5 cos 5x

T1 1.y=60x23-x +25 + y'=4Oy1t3-1T12. f(x) = cos (sin5 7x) +

f'(x) = -si6 (sins zx) . 5 sina 7x. cosTx .7= -35 sin (sins 7x) sinaTx cos 7x

T13. y'= 0.6 . .

(Function is y = -3 + 1.5x, for which the numericalderivative is 0.6081... .)

T14.v=3+5rl 6

v = Y'= -8x-2'6d = V'= 20.8f3'6

T15. f'(x) -72vsta + f(x) =gzxs4T16. f'(x) = 5 sin x and f(0) = 13

f(x)=-5cosx+C13=-5cos0+C =+ C= 18f(x)=-5cosx+18

T17. Carbon Dioxide ProblemD'

c(t)=300+Zcosffit

a. c'(t) =-ffi"i"fr,b. c'(273) =-#sin 1ft.zzs1

= 0.03442... ppm/day

c. Rate "#ffi, " m= 23e0.66...,

which is approximately 2390 tons per second!

44 Colculus: Concepts ond Applicotions Problem Set 3-10