I'm here again with another exercise (after this one) on Taylor series: determine the infinitesimal order for $x\to 0$ of the function $$ f(x)=\sqrt{\frac{x^4+x^5}{x-\sin{x}}}\ . $$ The problem is that, after having developed the denominator as $\frac{x^3}{6}+o(x^3)$, I have difficulties in determining the Taylor series of the whole function, as $f$ is not differentiable in $x=0$.
Thank you for your help!
Using steps $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^{9}\right)$$ $$x-\sin(x)=\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040}+O\left(x^{9}\right)$$ $$\frac{x^4+x^5}{x-\sin{x}}=\frac{x^4+x^5}{\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040}+O\left(x^{11}\right)}=6 x+6 x^2+\frac{3 x^3}{10}+O\left(x^{4}\right)$$ $$\sqrt{\frac{x^4+x^5}{x-\sin{x}}}=\sqrt{6} \sqrt{x}+\sqrt{\frac{3}{2}} x^{3/2}+O\left(x^{5/2}\right)$$