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asambeis [7]
3 years ago
10

Match each value with the smallest number set that it belongs to.

Mathematics
1 answer:
bulgar [2K]3 years ago
4 0
(-2)^3
12^0
I don't know the rest
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n^2-10n+22=-2
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Lim x-1 x2 - 1/ sin(x-2)
balu736 [363]

Answer:

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

Explanation:

Assuming the correct expression is to find the following limit:

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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:

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Evaluate the denominator:

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

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

4 0
3 years ago
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