<s>[INST] On understanding the use of binomial theorem to find asymptotes of a real valued function. I have the function 
$$y=\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$$
Wolfram Alpha says that it has a non linear asymptote at
$y=\frac{x^2}{2} - \frac{3x}{16}+ \frac{251}{256}$. 
I have been told to expand using the binomial theorem to obtain the asymptote at predicting-non-linear-asymptotes-of-a-real-valued-function ; however, I do not understand this. How does expanding binomially give me the asymptote ?
 [/INST] Consider first the denominator of $$y=\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$$ $${\sqrt{4x^2+3x+2}}=2x{\sqrt{1+\frac{3x}{4x^2}+\frac2{4x^2}}}=2x{\sqrt{1+\frac{3}{4x}+\frac1{2x^2}}}$$ Now apply the binomial therorem or Taylor series to get
$${\sqrt{1+\frac{3}{4x}+\frac1{2x^2}}}=1+\frac{3}{8 x}+\frac{23}{128 x^2}+O\left(\frac{1}{x^3}\right)$$ $$y=\frac 1 {2x}\frac{x^3+2x+9}{1+\frac{3}{8 x}+\frac{23}{128 x^2}+O\left(\frac{1}{x^3}\right)}$$ Now, use the long division to get $$y=\frac{x^2}{2}-\frac{3
   x}{16}+\frac{251}{256}+O\left(\frac{1}{x}\right)$$
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