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across

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We're here to help.

alfira

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alfira

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find the horzontal asymptotes

across

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All you have to do is determine the behavior of \(f\) as it approaches both \(-\infty\) and \(\infty\), i.e.,\[\lim_{x\to\infty}\frac{x^2+3}{\sqrt{x^2+1}}=?\]

alfira

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equal to 1 ?

alfira

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y=1?

across

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How did you get that?

alfira

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x^2 / X^2 +3/x^2

alfira

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everything divided by x^2

alfira

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because highest power in the function is x^2

across

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Recall that\[\lim_{x\to\infty}\frac{x^2+3}{\sqrt{x^2+1}}=\lim_{x\to\infty}\frac{x^2}{\sqrt{x^2}}=\lim_{x\to\infty}\frac{x^2}{|x|}=\lim_{x\to\infty}x=\infty.\]

across

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So it has no horizontal asymptote as \(x\to\infty\). What about \(x\to-\infty\)?

alfira

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same thing ?

across

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Right.

alfira

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thanks `