The Real And Complex Number Systems

f k x f k 0
x 0
fk x
x by induction hypothesis
P
e1/x2 3k 1/x
x
tP 3kt
(let t 1/x)
e t2
tP 3k t
e t
e t
e t2 0ast ( x 0) by (*).
Hence, f k1 0 0. So, by Mathematical Induction, we have proved that f n 0 0,
(n 1,2,...).
Since
lim
x0
f n x lim
x0
e 1/x2 P 3n 1/x
lim
x0
lim t
P 3n 1/x
e 1/x2
P 3n t
e t
0by(*)
f n 0,
we know that f n x is continuous at 0.
Remark: 1. Here is a proof on (*). Let Px be a real polynomial of degree n, and
choose an even number 2N n. We consider a Taylor Expansion with Remainder as
follows. Since for any x, wehave
2N1
e x 1k! x k e x,0
2N1
2N 2! x2N2 1k! x k ,
k0
k0
then
0 Px Px
e x
0asx
2N1 1
k! xk
k0
2N1
since degPx n deg 1
k0 k! xk 2N 1. By Sandwich Theorem, wehave
proved
Px
lim
x e x 0.
2. Here is another proof on f n 0 0, (n 1,2,...). By Exercise 5.15, it suffices to
show that lim x0 f n x 0. For the part, we have proved in this exercise. So, we omit the
proof. Exercise 5.15 tells us that we need not make sure that the derivative of f at 0. The
reader should compare with Exercise 5.15 and Exercise 5.5.
3. In the future, we will encounter the exercise in Charpter 9. The Exercises tells us one
important thing that the Taylor’s series about 0 generated by f converges everywhere
on R, but it represents f only at the origin.
5.5 Define f, g, andh as follows: f0 g0 h0 0 and, if x 0,
fx sin1/x, gx x sin1/x, hx x 2 sin1/x. Show that
(a) f x 1/x 2 cos1/x, ifx 0; f 0 does not exist.
e t
e t2