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Ratling [72]
3 years ago
14

Lim x-1 x2 - 1/ sin(x-2)

Mathematics
1 answer:
balu736 [363]3 years ago
4 0

Answer:

           \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0

Explanation:

Assuming the correct expression is to find the following limit:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}

Use the property the limit of the quotient is the quotient of the limits:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}

Evaluate the numerator:

          \frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}

Evaluate the denominator:

  • Since         \lim_{x \to1}sin(x-2)\neq 0

                  \frac{0}{\lim_{x \to1}sin(x-2)}=0

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Answer:

a: 3

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3 0
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raketka [301]
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7 0
3 years ago
The volume of a rectangular prism is represented by the function x3 + 11x2 + 20x – 32. The width of the box is x – 1 while the h
Oksi-84 [34.3K]

we know that

the volume of a rectangular prism is equal to

V=L*W*H

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W is the width of the box

H is is the height of the box

in this problem we have

V=x^{3\ }+11x^2+20x-32

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H=x+8

L=?

<u>Find the length of the box</u>

Using a graph tool-------> we will determine the roots of the equation of volume

see the attached figure

the roots are

x=-8\\x=-4\\ x=1

so

x^{3\ }+11x^2+20x-32=(x+8)*(x+4)*(x-1)

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<u>the answer is</u>

the length of the box is equal to (x+4)

8 0
3 years ago
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3 years ago
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Answer:

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7 0
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