Page 86 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 4] DERIVATIVES 77

1 1
Since lim f ðxÞ¼ lim cos þ 2x sin does not exist (because lim cos 1=x does not exist), f ðxÞ
0
0
x!0 x!0 x x x!0
cannot be continuous at x ¼ 0in spite of the fact that f ð0Þ exists.
0
This shows that we cannot calculate f ð0Þ in this case by simply calculating f ðxÞ and putting x ¼ 0,
0
0
as is frequently supposed in elementary calculus. It is only when the derivative of a function is
continuous at a point that this procedure gives the right answer. This happens to be true for most
functions arising in elementary calculus.
4.6. Present an ‘‘ ; ’’ definition of the derivative of f ðxÞ at x ¼ x 0 .
f ðxÞ has a derivative f ðx 0 Þ at x ¼ x 0 if, given any > 0, we can find > 0suchthat
0

f ðx 0 Þ < when 0 < jhj <

f ðx 0 þ hÞ f ðx 0 Þ
0

h

RIGHT- AND LEFT-HAND DERIVATIVES
4.7. Let f ðxÞ¼jxj.(a) Calculate the right-hand derivatives of f ðxÞ at x ¼ 0. (b) Calculate the left-
hand derivative of f ðxÞ at x ¼ 0. (c) Does f ðxÞ have a derivative at x ¼ 0? (d) Illustrate the
conclusions in (a), (b), and (c) from a graph.
y
jhj 0 h
f þ ð0Þ¼ lim f ðhÞ f ð0Þ ¼ lim ¼ lim ¼ 1
0
ðaÞ
h h h!0þ h
h!0þ h!0þ
since jhj¼ h for h > 0. y = _ y = x
x
jhj 0 h
f ð0Þ¼ lim f ðhÞ f ð0Þ ¼ lim ¼ lim ¼ 1
0
h h h!0 h
ðbÞ
h!0 h!0 x
since jhj¼ h for h < 0.
Fig. 4-8
(c) No. The derivative at 0 does not exist if the right and
left hand derivatives are unequal.
(d) The required graph is shown in the adjoining Fig. 4-8.
Note that the slopes of the lines y ¼ x and y ¼ x are 1 and 1 respectively, representing the right and
left hand derivatives at x ¼ 0. However, the derivative at x ¼ 0 does not exist.
2
4.8. Prove that f ðxÞ¼ x is differentiable in 0 @ x @ 1.
Let x 0 be any value such that 0 < x 0 < 1. Then
2 2
ðx 0 þ hÞ x 0
f ðx 0 Þ¼ lim ¼ lim ¼ limð2x 0 þ hÞ¼ 2x 0
f ðx 0 þ hÞ f ðx 0 Þ
0
h!0 h h!0 h h!0
At the end point x ¼ 0,
2
h 0
f þ ð0Þ¼ lim f ð0 þ hÞ f ð0Þ ¼ lim ¼ lim h ¼ 0
0
h h
h!0þ h!0þ h!0þ
At the end point x ¼ 1,
2
ð1 þ hÞ 1
f ð1Þ¼ lim f ð1 þ hÞ f ð1Þ ¼ lim ¼ lim ð2 þ hÞ¼ 2
0
h h
h!0 h!0 h!0
Then f ðxÞ is differentiable in 0 @ x @ 1. We may write f ðxÞ¼ 2x for any x in this interval. It is
0
customary to write f þ ð0Þ¼ f ð0Þ and f ð1Þ¼ f ð1Þ in this case.
0
0
0
0