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alex41 [277]
3 years ago
8

PLEASE PLEASE PLEASE HELP ILL GIVE BRAINEST PLEAESE​

Mathematics
2 answers:
VMariaS [17]3 years ago
6 0
The answer is D btw because there are two of the same x values
AveGali [126]3 years ago
5 0

Answer:

The last one

Step-by-step explanation:

There are two of the same x-values, meaning it is not a function.

Hopefully this helps!

Brainliest please?

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<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\

----------

Part II

\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\

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Part III

\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\

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