Finding the oblique asymptote of:

Question

Given $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}$$ how would you find the oblique asymptote of that?

Answer 1

When $x$ is large, $x^2+1 \approx x^2$ and $\sqrt{x+1} \approx \sqrt{x}$. So, for large values of $x$, $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}\approx x^{3/2}$$ I suppose that you consider "oblique" in a very extended way.

More precise answers could be obtained writing $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}=\frac{x^2+1}{\sqrt{x} \sqrt{1+\frac1x}}$$ and considering the Taylor expansion $$\frac{1}{\sqrt{1+y}}=1-\frac{y}{2}+\frac{3 y^2}{8}+O\left(y^3\right)$$ Replace $y$ by $\frac{1}{x}$ and obtain, for large values of $x$ $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}=x^{3/2}-\frac{\sqrt{x}}{2}+\frac{11 \sqrt{\frac{1}{x}}}{8}-\frac{13}{16} \left(\frac{1}{x}\right)^{3/2}+O\left(\left(\frac{1}{x}\right)^{5/2}\right)$$ So the asymptotic curve is $$g(x)=x^{3/2}-\frac{\sqrt{x}}{2}$$ and it is approached from above when $x$ tends to $\infty$.