Asymptotic formula for cot(x).

Question

So I'm asked to pick such $A$ and $B$ that when $x$ approaches $0,$

$$\cot(x) = (1 + Ax^2)/(x + Bx^2) + O(x^5)$$1

I have no idea how to do that. Any help appreciated!

Answer 1

The expression in the image seems to be $$ \text{ctg}\,x=\frac{1+Ax^2}{x+Bx^{\color{red}{3}}}+O(x^5). \tag{1} $$ The series expansion at $x=0$ of the LHS of $(1)$ gives $$ \text{ctg}\,x=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}+O(x^5), \tag{2} $$ whereas for the RHS it yields \begin{align} \frac{1+Ax^2}{x+Bx^3}&=\frac{1}{x}(1+Ax^2)(1-Bx^2+B^2x^4+O(x^6)) \\ &=\frac{1}{x}+(A-B)x+(B^2-AB)x^3+O(x^5). \tag{3} \end{align} Equating coefficients of equal powers of $x$ we obtain the system of equations $$ \begin{cases} A-B=-\frac{1}{3}, \\ B^2-AB=-\frac{1}{45}, \end{cases} \tag{4} $$ the solution of which is $$ (A,B)=\left(-\frac{2}{5},-\frac{1}{15}\right). \tag{5} $$