# Can anybody help me finding asymptotes of a function?

The function given is $f(x) = \sqrt{{x}^2(2-x)}$.

Can anybody help me to find all asymptotes of this function. I know it doesn't have a vertical asymptote and I know that it's horizontal asymptote is $\sqrt{-1}$, but I don't know how to find asymptote of the slope.

I'd prefer if someone could help me solving it using the formula given below: $y = kx + l$ where $k = lim_{n\to\infty} \dfrac{f(x)}{x}$ and $l=lim_{n\to\infty}[f(x)-kx]$. I found $k$ that is $k=-1$ but I don't know how to find $l$.

You want to compute

$$k=\lim_{x\to\infty}\left(\sqrt{{x}^2(2-x)}+x\right).$$

To get rid of the cubic root, you can multiply by the conjugate trinomial and get

$$k=\lim_{x\to\infty}\left(\frac{{x}^2(2-x)+x^3}{\sqrt{{x}^2(2-x)}^2-\sqrt{{x}^2(2-x)}x+x^2}\right).$$

The numerator simplifies to $2x^2$ and by factoring out $x^2$ at the denominator, the expression tends to

$$\frac2{(-1)^2-(-1)+1}.$$

$\sqrt{x^2(2-x)}=-x\sqrt{1-\frac2x}=\boxed{\text{via Taylor}}=-x(1-\frac2{3x}+o(\frac1x))=-x+\frac23+o(1)$

As $\sqrt{x^2(2-x)}- (-x+\frac23)$ tends to $0$ the asymptotes at $-\infty$ and $+\infty$ are $y=-x+\frac23$ according to definition.

Alternative approach is standart: firstly search $k=\lim\limits_{x\to\pm\infty}\frac {f(x)}{x}$ and secondly define $b=\lim\limits_{x\to\pm\infty}(f(x)-kx)$. If both calculations would successful then the asymptotes would be $y=kx+b.$ They may be different at $-\infty$ and $+\infty$ or exist only at one of the infinities.

• To find b,we use the following formula $b = lim_{n\to\infty}(f(x)-kx)$, so in your case you didn't solve it by this formula, or you did but didn't show me how to. Could you please solve it again? I found that k = -1 but I don't know how to find b. If there is another way than Taylor, please show it to me. – Vala Ahmeti Apr 21 '17 at 13:25
• @Vala Ahmeti, Just calculate $b=\lim\limits_{x\to\infty}(\sqrt{x^2(2-x)}+x)=\lim\limits_{x\to\infty}\dfrac{(\sqrt{\hspace{.2cm}})^3+x^3}{(\sqrt{\hspace{.2cm}})^2-x(\sqrt{\hspace{.2cm}})+x^2}=\ldots$ – Minz Apr 22 '17 at 1:35