2 years ago

A,b,c or d please help i don’t understand this

1 answer:

7 0

uhm the answer is D. sorry if am wrong

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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:

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

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:

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

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Step-by-step explanation:

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4^8 is the answer according to the law of indices

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