Oblique asymptote of $f(x) = \frac{x\sqrt{x}}{\sqrt{x+2}}$

Question

I've found the slope using $\lim_{x \to \infty } \frac{f(x)}{x}$ that is $1$ but I can't find the $h = \lim_{x \to \infty } f(x) - x$ . How we find $\lim_{x \to \infty}\frac{x\sqrt{x}}{\sqrt{x+2}} - x$ if it exists ?

Also if there is another way for finding oblique asymptote , please write it .

Answer 1

$$\lim_{x\rightarrow+\infty}\left(\frac{x\sqrt{x}}{\sqrt{x+2}} - x\right)=-\lim_{x\rightarrow+\infty}\frac{2x}{(\sqrt{x+2}+\sqrt{x})\sqrt{x+2}}=$$ $$=-\lim_{x\rightarrow+\infty}\frac{2}{\left(\sqrt{1+\frac{2}{x}}+1\right)\sqrt{1+\frac{2}{x}}}=-1,$$ which gives $y=x-1$.

Answer 2

$$\frac{x\sqrt{x}}{\sqrt{x+2}} - x=\frac{x\sqrt{x}-x\sqrt{x+2}}{\sqrt{x+2}}\frac{x\sqrt{x}+x\sqrt{x+2}}{x\sqrt{x}+x\sqrt{x+2}}=\frac{x^3-x^3-2x^2}{\sqrt{x+2}(x\sqrt{x}+x\sqrt{x+2})}=\frac{-2x^2}{x\sqrt{x^2+2x}+x\sqrt{x^2+4x+4}}=\frac{-2x^2}{x^2\left(\sqrt{1+\frac2x}+\sqrt{1+\frac4x+\frac4{x^2}}\right)}\to\frac{-2}{2}=-1$$

Answer 3

Alternative way. By using the Taylor expansion $(1+t)^a=1+at+o(t)$ as $t\to 0$, we can find the oblique asymptote in just one step: as $x\to +\infty$ $$f(x)=\frac{x\sqrt{x}}{\sqrt{x+2}}=x\left(1+\frac{2}{x}\right)^{-1/2}=x\left(1-\frac{1}{2}\cdot\frac{2}{x}+o(1/x)\right)=x-1+o(1).$$ Hence $\lim_{x\to +\infty}(f(x)-(x-1))=0$, which means that the oblique asymptote is $y=x-1$.