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?
Determine whether the lines
L₁ (t) = (-2,3, −1)t + (0,2,-3) and
L2 p(s) = (2, −3, 1)s + (-10, 17, -8)
intersect. If they do, find the point of intersection.
Convert the line given by the parametric equations y(t)
Enter the symmetric equations in alphabetic order.
(x(t)
= -4+6t
= 3-t
(z(t)
=
5-7t
to symmetric equations.
Find the point at which the line (t) = (4, -5,-4)+t(-2, -1,5) intersects the xy plane.
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