On understanding the use of binomial theorem to find asymptotes of a real valued function.

Question

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 ?

Answer 1

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 $$y=\frac{x^2}{2}-\frac{3 x}{16}+\frac{251}{256}+O\left(\frac{1}{x}\right)$$

Answer 2

hint

Let $$f (x)=(x^3+2x+9)(4x^2+3x+2)^\frac{-1}{2}$$

$$g (x)=f (1/x)=$$ $$(1+\frac {9x}{2}+\frac {1}{2x^2})(1+\frac {3x}{4}+\frac {x^2}{2})^\frac {-1}{2} $$

expand $g (x) $ near $x=0$ using binomial formula.

$$(1+\frac {3x}{4}+\frac {x^2}{2})^\frac {-1}{2} $$ $$=1-\frac {3x}{8}-\frac {x^2}{4}+\frac {27x^2}{112}+x^2\epsilon (x)$$

You will get $f (x)=g (1/x) $ and the asymptote.