3 years ago

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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It's almost solved. Just move the -2 to the other side.
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balu736 [363]

Answer:

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

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

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

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