This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Page 795 I1 + 2Iis less than 5. Appendix A
APPENDIX A
Introduction to the Theory of Limits
1. For any chosen value of 6 we want the As we approach 2, 4. distance between f(x) and L to be less than 6.
In absolute value notation this is {f(sc) — LI<E
For our function and limit we need 6.
[(22 — 5) — (—3)<6
l2zr —2I <6
2:c — 1< 6 So the distance between z and 1 needs to be
less than 3. Therefore, let 6 = %. I31+7 —1]=3lz+2l<6sothat
6 IZ+2I < '3
Pick6=§
lim (32+ 1): 4 gé 5; the interval on the y
1—»1 axis in the neighborhood corresponding to
x = 1 can be arbitrarily restrictive. 0<z1<6 must apply and the given statement is false.
The actual limit is 4.
lim (22 — 2)=5im_plieslz2 — 2 — 5l<e. 9. H2 Forlz— 2I<6,2 — 6<27<2+6. The doubter picks 6 = 1. The 6 inequality 10 becomes
—1<12  7<1,6<£2<80r 2.45 < «E < 2.83. The believer has the
impossible task of ﬁnding a number around
a: = 2. The given statement is false. The actual limit is 2. For any chosen value of 6 we want the
distance between f (z) and L to be less than 6. In absolute value notation this is 12 ”(2) — L<6 For our function and limit we need 13 l(z2 + 2) — (6)l<6
'22 — 4i<€ lz— 2llz+ 2l<e 11. If the distance between z and 2, written as
I2 — 2 , is to be less than 6, we need
In: — 2K5) < 6 = 56 . Therefore, let 6 = minimum of{1,§ .
'12 — 32+2l=(z —2)(:c —1)
Letlz — llbe less than the minimum of
%or6. Then%ng%andlx — 2I_<_%.
Thus, Iz— 2IIz— 1 < g6=€0r6=32§€ The statement is true; no matter what
number the doubter selects, the believer will
always be able to ﬁnd '22 — 3x + 2l< 6. [f(x) — L=[(z+3) — 5I=iz — 2<6
Choose 6 = 6. f(t) — L=(3i — 1) — 0=31— 1 _ 1  3 * 3 The statement is false. Choose 6 = 0.3 and it
is not possible to ﬁnd a delta, because lt—§]<§=0.1 f(z)  L=(3z + 7) — 1=3z + 6
=3z+2l<36;choose6 2%. [f(r) —— Ll=[(2r  5)+3:2iz — 1
< 26; choose6=%. [f(z) — Ll=l(22 + 2) — 6=z2 — 41
=lr — 2H1 + 2; choose6=min(1, g)
aslz—2l<6andz—2I<1means
—1<z<3or1<z+2<5so
lz—2llz+2<6(5)=6. mm) — LI: 1 1—”’——2" _ T ' 2m ’
choose 6 = min(1, é)‘ 1' 2'. In order for f(z‘) to be continuous at a: : 0 , yin) f(z) must equal f(0). To show that lim sin 3.13
2—40 any 6 > 0 there exists a 6 > 0 such that = 0 we need to show that for Page 796 14. 15. f(z)  Ll < c whenlx — 0 < 6. 16. Arbitrarily letting c = 0.5 we need to find a
6interval about 0 such that lsin %l < 0.5. However any interval about 0 contains a point 1: = % (with n odd) where sin % 7rn
= sm— =
2 :l: 1. Therefore, there does not
exist a 6interval about 0 such that lsin :15! < 0.5, and f(z) must be
discontinuous at :c = 0. By hypothesis, lime f(2:) = L which means that given 61 > 0 such that f(.7:) — LI< 61 there exists 6 for 0 <12 — cl < 6. Let 17. 9(2) = kf(x) so that
19(2) — kLl=lkfm — kLl
:lklf(:c) — L<Ilce1 This says given 6 > 0, let 61 — L and then there isa 45 such that I“
lg(1:) — [CL] < c
if '
0<r — cl<6 18. This inequality is restated as lmcﬁz) 2 [CL or lime km) = M. Given 6 > 0 there exists 61 and 62 such that 19. f(a:) — L1l<gfor0<lz — c<61
and [9(1) — L2l<§for0<lz — c<62.
Then ”(1‘)" 9(3) — (L1 — L2” = Ilf(1‘)  Ll]  [9(1)  L2H 20
SIf(I)  L1l+lg($) — Lzl
<§+§=6 if0<z — cl < min(61, 62). 21. This says that 1:21;] f(.r) —— g(z) I = L1 — L2. Appendix A If lianfﬂt) = L1 and lli‘rgcﬂz) = L2
then given any 6 > 0, If(:c) — L1< 2%] 6 c<6 andg(:c) — L2I<Tbif0<lz — Hence, laf(z) + 69(3) — a(L1 + 6L2)
=lalf(z) “ L1] + I’M"3) — L2H
smilm) — LII+Ibugm — L2
<Ia12—fa—l+lbI—£=e 2 bl
This says that Iggclaﬂm bg(z>1 = a Ignore) + b 133,6
9(1)
Given 6 > 0,f(z) — 0< ‘ﬂfor 0<z — c<61
”(1') — 0l< \/Efor 0<l1: — c<62
So
mam  0mm — ongm — 0
< \/E \/E = c if0 <Iz  cl< min(61, 62)
This says that lmcf(:c)g(z) = 0.
05lg(x) — Olglﬂz) — 0l<c whenever
0<z — c<6. Thus,g(z) — 0l<c for
0 <11 — cl<6 implies 1330 9(2) = 0.
Let MI) = f(I)  9(1) Since ﬂat) 2 9(2)
throughout an open interval containing c, the
limit limitation theorem guarantees that
Lilac be) = aggro) — gm] = lime we) — Iggc 9(1) 2 0 or
Imam 2 1}an gm.
0: Hf(x)l — lLllslf(:v) — L<e
whenever 0 <z — c< 6. Thus,
“f(a:) — lLl l< 6 implies llilnﬂlﬂx)! = ILI.
By hypothesis, 1:3 c f (z) = L, and by Problem 20,1;gcf(x)=ILl, or ifs = Iz—L—l> 0 then
there exists 6 so that “f(:c)' —Ll' < lél for all 6 such that 0 < I z — cl < 6. Thus, —ITLI<(f(z)[ — ILI)<'7“and AppendixA
l—g—<f(ac)<?’l2£l.
22. a. lf(z)<i2L—l,
mz) )+L<f(z)l+Ll<Ll+3l——2Ll=5ITM by Problem 21, if0 <Ia: — c < 61.
b. Given 62 > 0, there exists 6 such that
[f(x) — Ll<62if0<z — “<6 Then
[[f (as)? — L2l=lf(z)+ Lllf(z) — LI
<5—l2£€2 ifO <12: — c<min(61, 6) = 62
c. Let 6 .. 5—l—2Llc2, thenl[f( [f(]:r:)2 — L2l< c
for 0 <  2— cl < 62 where 62: min(61, 6)
23 i—[U + g)2  (f  9)]
=%[f2+2f9+ 92 — f2+2fg — f]
= Hm + 2ng
= I!)
Thus,
Iggy: Iggjw + 9)? — (f — gﬂ = £1,311.40 + g)?  (f — m
by Problem 14. By Problem 15, it also equals
1  2 . 2 Iggy + a) — Iggy — w]
by Problem 22 it becomes 3.1[(L+M)2 — (L — M)2]=LM
24. Letc=%c).r_’ lim f(z): (c) implies
If ()— f(c>l< —c)
for0<lz — Cl<6.
—f——(—c)<f(r) — f(C)< 52—”)
o < {(2—62 < f(z) < 3%; ,0<f(z) for0<l17  cl<6.
25. 3.. Since f is continuous at L, we have
lim f (w) = f(L) which means that there
w—’L
exists a 61 such that [f(w) — f(L) I < ‘1 Wheneverlw — LI< 61. Page 797 b. Since
19(2) — LI < 62 lime g(z) = L, by hypothesis, whenever 0 <:r — c< 62.
Let w = g(z) in part a, and c2 = 61.
Then lf[g(z)l —f(L)l < 61 whenever 0 <  2— cl < 62. 26. Since (as it turns out) the slope at z = 3 is
— 26; pick K larger than 26 works. ...
View
Full Document
 Spring '08
 all

Click to edit the document details