Hi i’m not sure how to do question 20 if u could explain how to do it that’d b great !!
1 answer:
Answer:
A) -2
Step-by-step explanation:
The form is indeterminate at x=0, so L'Hopital's rule applies. The resulting form is also indeterminate at x=0, so a second application is required.
Let f(x) = x·sin(x); g(x) = cos(x) -1
Then f'(x) = sin(x) +x·cos(x), and g'(x) = -sin(x).
We still have f'(0)/g'(0) = 0/0 . . . . . indeterminate.
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Differentiating numerator and denominator a second time gives ...
f''(x) = 2cos(x) -sin(x)
g''(x) = -cos(x)
Then f''(0)/g''(0) = 2/-1 = -2
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I like to start by graphing the expression to see if that is informative as to what the limit should be. The graph suggests the limit is -2, as we found.
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Answer:
Let the vectors be
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
c / | c |
Where | c | = √ (7)² + (2)² + (-1)² = 3√6
Therefore, the unit vector is
or
[ ,
,
]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[ ,
,
]
In conclusion, the two unit vectors are;
[ ,
,
]
and
[ ,
,
]
<em>Hope this helps!</em>
Answer:
Step-by-step explanation:
f'(x)=-sin x
c=0∈(-π/3,π/3)