One cannot put h = 0 in the formula f ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) 2 h for the derivative of f . (See Exercise 56.) However, in the last step of each of the computations in the text, we are effectively setting h = 0 whentaking the limit. What is going on here?
One cannot put h = 0 in the formula f ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) 2 h for the derivative of f . (See Exercise 56.) However, in the last step of each of the computations in the text, we are effectively setting h = 0 whentaking the limit. What is going on here?
Solution Summary: The author explains that the value h=0 should be substituted in the formula of derivative, based on the ordinary difference quotient formula.
f
′
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
2
h
for the derivative of f. (See Exercise 56.) However, in the last step of each of the computations in the text, we are effectively setting
h
=
0
whentaking the limit. What is going on here?
a
->
f(x) = f(x) = [x] show that whether f is continuous function or not(by using theorem)
Muslim_maths
Use Green's Theorem to evaluate F. dr, where
F = (√+4y, 2x + √√)
and C consists of the arc of the curve y = 4x - x² from (0,0) to (4,0) and the line segment from (4,0) to
(0,0).
Evaluate
F. dr where F(x, y, z) = (2yz cos(xyz), 2xzcos(xyz), 2xy cos(xyz)) and C is the line
π 1
1
segment starting at the point (8,
'
and ending at the point (3,
2
3'6
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