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Find the asymtode of $(9x^3 + 5)/(2x^3+\sqrt{x^6+2})$

i know how to find the horizontal and oblique asymptotes but i am stuck with this question. how to evaluate the square root part as that too will give a $x^3$. and since the degree of the numerator and denominator are 3 do we need to take into consideration the $x^6$ part in root ?

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  • $\begingroup$ Hint: As $x$ approaches large values, what does $x^6+2$ approximate? $\endgroup$ – Sherlock Holmes Oct 23 '14 at 23:25
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If $x\to \infty$ then

$$\frac{9x^3+5}{2x^3+\sqrt{x^6+2}}=\frac{\frac{9x^3}{x^3}+\frac{5}{x^3}}{\frac{2x^3}{x^3}+\frac{\sqrt{x^6+2}}{x^3}}=\frac{9+\frac{5}{x^3}}{2+\sqrt{\frac{x^6}{x^6}+\frac{2}{x^6}}}=\frac{9+\frac{5}{x^3}}{2+\sqrt{1+\frac{2}{x^6}}}\to \frac{9}{3}=3$$

If $x\to -\infty$ then

$$\frac{9x^3+5}{2x^3+\sqrt{x^6+2}}=\frac{\frac{9x^3}{x^3}+\frac{5}{x^3}}{\frac{2x^3}{x^3}+\frac{\sqrt{x^6+2}}{x^3}}=\frac{9+\frac{5}{x^3}}{2-\sqrt{\frac{x^6}{x^6}+\frac{2}{x^6}}}=\frac{9+\frac{5}{x^3}}{2-\sqrt{1+\frac{2}{x^6}}}\to \frac{9}{1}=9$$

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As $x$ approaches large values, $x^6+2 \approx x^6$. You can use this to find the asymptotes.

$\frac{9x^3+5}{2x^3+\sqrt{x^6}}=\frac{9x^3+5}{3x^3}=3+\frac{5}{3x^3}$

So an asymptote is $y=3$.

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  • $\begingroup$ oh that makes sense, thanks $\endgroup$ – sangam.saga Oct 23 '14 at 23:25

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