Page 85 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 85
76 DERIVATIVES [CHAP. 4
1
f ðx þ hÞ f ðxÞ 3 3
ðaÞ ¼ ½ðx þ hÞ x
h h
1
3
2
2
3
3
2
¼ ½x þ 3x h þ 3xh þ h x ¼ 3x þ 3xh þ h 2
h
Then
f ðxÞ¼ lim f ðx þ hÞ f ðxÞ ¼ 3x 2
0
h!0 h
p ffiffiffiffiffiffiffiffiffiffiffi
p ffiffiffi
x
lim f ðx þ hÞ f ðxÞ ¼ lim x þ h
ðbÞ
h!0 h h!0 h
ffiffiffiffiffiffiffiffiffiffiffi
p p ffiffiffi
The result follows by multiplying numerator and denominator by x þ h x and then letting h ! 0.
4.3. If f ðxÞ has a derivative at x ¼ x 0 , prove that f ðxÞ must be continuous at x ¼ x 0 .
h; h 6¼ 0
f ðx 0 þ hÞ f ðx 0 Þ
f ðx 0 þ hÞ f ðx 0 Þ¼
h
Then lim f ðx 0 þ hÞ f ðx 0 Þ¼ lim f ðx 0 þ hÞ f ðx 0 Þ lim h ¼ f ðx 0 Þ 0 ¼ 0
0
h!0 h!0 h h!0
since f ðx 0 Þ exists by hypothesis. Thus
0
lim f ðx 0 þ hÞ f ðx 0 Þ¼ 0 or lim f ðx 0 þ hÞ¼ f ðx 0 Þ
h!0 h!0
showing that f ðxÞ is continuous at x ¼ x 0 .
x sin 1=x; x 6¼ 0
0; x ¼ 0
4.4. Let f ðxÞ¼ .
(a)Is f ðxÞ continuous at x ¼ 0? (b) Does f ðxÞ have a derivative at x ¼ 0?
(a)By Problem 3.22(b)of Chapter 3, f ðxÞ is continuous at x ¼ 0.
h sin 1=h 0 1
f ð0Þ¼ lim f ð0 þ hÞ f ð0Þ ¼ lim f ðhÞ f ð0Þ ¼ lim ¼ lim sin
0
ðbÞ
h!0 h h!0 h h!0 h h!0 h
which does not exist.
This example shows that even though a function is continuous at a point, it need not have a
derivative at the point, i.e., the converse of the theorem in Problem 4.3 is not necessarily true.
It is possible to construct a function which is continuous at every point of an interval but has a
derivative nowhere.
2
x sin 1=x; x 6¼ 0
0; x ¼ 0
4.5. Let f ðxÞ¼ .
(a)Is f ðxÞ differentiable at x ¼ 0? (b)Is f ðxÞ continuous at x ¼ 0?
0
2
h sin 1=h 0 1
f ð0Þ¼ lim f ðhÞ f ð0Þ ¼ lim ¼ lim h sin ¼ 0
0
ðaÞ
h!0 h h!0 h h!0 h
by Problem 3.13, Chapter 3. Then f ðxÞ has a derivative (is differentiable) at x ¼ 0 and its value is 0.
(b)From elementary calculus differentiation rules, if x 6¼ 0,
d 1 2 d 1 1 d
2
0 x sin ¼ x sin þ sin 2
dx x dx x x dx
f ðxÞ¼ ðx Þ
1 1 1 1 1
¼ x 2 cos þ sin ð2xÞ¼ cos þ 2x sin
x x 2 x x x
