Find the asymptote of the function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$

Question

We have a function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ and when $x$ goes towards $-\infty$, we have an asymptote $y = -2x - 3/2$. How we get this asymptote?

Answer 1

You can apply simple algebraic transformations, being careful to always have positive expressions under the square root. For $x<0$, and thus $|x|=-x$, the expression is equal to \begin{align} =\frac{-x\sqrt{-x}}{\sqrt{3-x}}-x &=\frac{x}{\sqrt{3-x}}(-\sqrt{-x}-\sqrt{3-x})\\ \\ &=-2x+\frac{x}{\sqrt{3-x}}(-\sqrt{-x}+\sqrt{3-x}) \\ &=-2x-\frac{-x}{\sqrt{3-x}}\frac{3}{\sqrt{-x}+\sqrt{3-x}} =-2x-\frac{3}{\sqrt{1-\frac3x}+1-\frac3x} \end{align} which asymptotically is equal to $-2x-\frac32$ for $x\to-\infty$.

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For $x>3$ the expression is equal to $$ \frac{x}{\sqrt{x-3}}(\sqrt{x}-\sqrt{x-3}) =\frac{x}{\sqrt{x-3}}\frac{x-(x-3)}{\sqrt{x}+\sqrt{x-3}} =\frac{3}{\sqrt{1-\frac3x}+1-\frac3x} $$ which asymptotically converges to the constant $\frac32$.

Answer 2

$f(x)=\sqrt{\frac{x^3}{x-3}}-x$

$\lim_{x\to -\infty}\frac{f(x)}{x}=-2$

and

$\lim_{x\to -\infty}f(x)-(-2x)=-\frac{3}{2}$

Answer 3

Note that if $ax+b$ is an asymptote for the function $f(x)$ in $-\infty$, then$$a=\lim_{x\to -\infty}{f(x)\over x}\\b=\lim_{x\to -\infty}{f(x)-ax}$$therefore$$a{=\lim_{x\to -\infty}{\sqrt{x^3\over x-3}-x\over x}\\=\lim_{x\to -\infty}{-\sqrt{x\over x-3}-1}\\=-2}$$and $$b{=\lim_{x\to -\infty}\sqrt{x^3\over x-3}-x+2x\\=\lim_{x\to -\infty}\sqrt{x^3\over x-3}+x\\=\lim_{x\to -\infty}{{x^3\over x-3}-x^2\over \sqrt{x^3\over x-3}-x}\\=\lim_{x\to -\infty}{{3x^2\over x-3}\over \sqrt{x^3\over x-3}-x}\\=\lim_{x\to -\infty}{3x^2\over -\sqrt{x^3(x-3)}-x(x-3)}\\=\lim_{x\to -\infty}{3\over -\sqrt{x-3\over x}-{x\over x-3}}\\=-{3\over 2}}$$

Answer 4

Let $$ f(x) =\sqrt{\frac{x^3}{x-3}}-x =|x|\sqrt{\frac{x}{x-3}}-x $$ For $x<0$ we have $|x|=-x$, thus $$ f(x) =-x\sqrt{\frac{x}{x-3}}-x =\Bigl(-\sqrt{\frac{x}{x-3}}-1\Bigr)x =\Bigl(-\sqrt{\frac{1}{1-3/x}}-1\Bigr)x. $$

For $x<0$; $$\lim_{x\to-\infty}\frac{f(x)}{x} =\lim_{x\to-\infty}\Bigl(-\sqrt{\frac{1}{1-3/x}}-1\Bigr) =-1-1=-2. $$

Now, study \begin{align*} f(x)-(-2x) & =f(x)+2x =-x\sqrt{\frac{x}{x-3}}-x+2x =-x\sqrt{\frac{x}{x-3}}+x =\Bigl(-\sqrt{\frac{x}{x-3}}+1\Bigr)x. \end{align*}

Maclaurin expansion gives \begin{align*} \sqrt{\frac{x}{x-3}} &=\sqrt{\frac{x-3+3}{x-3}} =\sqrt{1+\frac{3}{x-3}} =1+\frac{1}{2}\cdot\frac{3}{x-3}+\Bigl(\frac{3}{x-3}\Bigr)^2L(1/x) \end{align*} and we have \begin{align*} \Bigl(-\sqrt{\frac{x}{x-3}}+1\Bigr)x &=\Bigl(-\Bigl(1+\frac{1}{2}\cdot\frac{3}{x-3}+\Bigl(\frac{3}{x-3}\Bigr)^2L_1(1/x)\Bigr)+1\Bigr)x \\&=\Bigl(-1-\frac{1}{2}\cdot\frac{3}{x-3}-\frac{1}{(x-3)^2}L_2(1/x)+1\Bigr)x \\&=\Bigl(-\frac{3}{2}\cdot\frac{1}{x-3}-\frac{1}{(x-3)^2}L_2(1/x)\Bigr)x \\&=-\frac{3}{2}\cdot\frac{x}{x-3}-\frac{x}{(x-3)^2}L_2(1/x) \\&=-\frac{3}{2}\cdot\frac{1}{1-3/x}-\frac{x}{(x-3)^2}L_2(1/x) \\&\to-\frac{3}{2} \end{align*} as $x\to-\infty$, where $L_{1,2}(1/x)$ is limited for large $x$.

Hence $$ \lim_{x\to-\infty}\bigl(f(x)+2x\bigr)=-\frac{3}{2} $$ showing that $$ f(x)\,\,\mathrm{"}=\mathrm{"}\,-2x-\frac{3}{2} $$ for large negative $x$.