d. f(0) = rim
!d:{Q
T7. Graph. f(x) =
h+0 xu
xsin2x
lim = x+0 lim = xt 0
i!!I
(1)
sln x x
xsin2x+sinx xstnx
Using TABLE for numerator, denominator, and quotient shows that the numerator goes to zero faster than the denominator. For instance, if x = 0.00'l
As x approaches 0, f(x) approaches 1. The squeeze theorem. lt states that:
,
. Quotient
1.1666... x 109 _ {a = 0.001 16... . = 9.999... x 107 ^ ^^r Thus, the limit appears to be zero. (The limit can be found algebraically to equal zero by I'Hospital's rule after students have studied Section
lf (1) g(x) < h(x) for all x in a neighborhood of c, (2) h(x) = l, 6n6 s(x) =
68.)
Chopter Test T1. See definition of limit in Chapter 2. T2. See definition of derivative in Section 32 or 3'4. T3. Prove that if f(x) = 3t', then f'(x) = llx3. Proof
:
*;,l.a"tire=,1'lrg# lim = h+0
= T
3x4 + 12x3h + 18x2h2 + 12xh3 + 3h4

3x4
h
lim (12x3+ 18x2h + 12xh2 + 3h3; = 12xs, Q.E.D.
. t(x) = cos
3x
+
t'(x) =
Jg
,lg
(3) f is a function for which g(x) < f(x) < h(x) for all x in that neighborhood 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. T1
5x) = 5 cos 5x
*q{sin
1.y=60x23x +25
+
y'=4Oy1t31
T12. f(x) = cos (sin5 7x) + f'(x) = si6 (sins zx) . 5 sina 7x. cosTx = 35 sin (sins 7x) sinaTx cos 7x
.7
T13. y'= 0.6 . . (Function is y = 3 + 1.5x, for which the numerical
3 sin 3x.
t'(5) = 3 sin 15 = 1.95086... Decreasing at 1.95... yunits per xunit.
derivative is 0.6081... .)
T5. Graph (below). lf you zoom in on the point where x = 5, the graph appears to get closer and closer to the tangent line. The name of this property is local
linearity.
T14.v=3+5rl
6
v = Y'= 8x2'6 d = V'= 20.8f3'6 T15. f'(x) 72vsta + f(x) =gzxs4 T16. f'(x) = 5 sin x and f(0) = 13
f(x)=5cosx+C 13=5cos0+C =+ C= 18 f(x)=5cosx+18
T17. Carbon Dioxide Problem D'
c(t)=300+Zcosffit T6. Amos substituted before differentiating instead ol after. Correct solution is f(x) = 7x
+
f'(x)
=7 +
f'(5) = 7.
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 310