The Real And Complex Number Systems

|fx| |fa| |L| 1
which contradicts to lim xb
fx .
Hence, lim xb
f x either fails to exist or is infinite.
5.11 Show that the formula in the Mean Value Theorem can be written as follows:
fx h fx
f
h
x h,
where 0 1.
Proof: (Mean Value Theorem) Letf and g be continuous on a, b and differentiable
on a, b. Then there exists a a, b such that fb fa f b a. Note that
a b a, where 0 1. So, we have proved the exercise.
Determine as a function of x and h, and keep x 0 fixed, and find lim h0 in each
case.
(a) fx x 2 .
Proof: Consider
fx h fx
h
which implies that
x h2 x 2
h
2x h 2x h f x h
1/2.
Hence, we know that lim h0 1/2.
(b) fx x 3 .
Proof: Consider
fx h fx
x h3 x 3
3x
h
h
2 3xh h 2 3x h 2 f x h
which implies that
3x 9x2 9xh 3h 2
3h
Since 0 1, we consider two cases. (i) x 0, (ii) x 0.
(i) As x 0, since
0 3x 9x2 9xh 3h 2
1,
3h
we have
3x 9x 2 9xh3h 2
3h
if h 0, and h is sufficiently close to 0,
3x 9x 2 9xh3h 2
if h 0, and h is sufficiently close to 0.
3h
Hence, we know that lim h0 1/2 by L-Hospital Rule.
(ii) As x 0, we have
3x 9x 2 9xh3h 2
3h
if h 0, and h is sufficiently close to 0,
3x 9x 2 9xh3h 2
if h 0, and h is sufficiently close to 0.
3h
Hence, we know that lim h0 1/2 by L-Hospital Rule.
From (i) and (ii), we know that as x 0, we have lim h0 1/2.