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Anna11 [10]
2 years ago
8

Can somebody pleaseee draw the line on these

Mathematics
1 answer:
tamaranim1 [39]2 years ago
4 0

Answer:

<h3>6) (-1,5) , (1,-5) </h3>

Step-by-step explanation:

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Which of these limits evaluate to 0?
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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\\\\\\

----------

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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