Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{\alpha \cot x}{x}+\frac{\beta }{x^2},& 0<|x|\le 1\\ \frac{1}{3},& x=0\end{array}$

If $f\left(x\right)$ is continuous at $x=0,$ then the value of ${\alpha }^2+{\beta }^2$ is

Answer 1

$\underset{x\to 0}{lim}f\left(x\right)=\frac{1}{3}\Rightarrow \underset{x\to 0}{lim}\frac{x\cdot \alpha \cot x+\beta }{x^2}=\frac{1}{3}\Rightarrow \underset{x\to 0}{lim}\frac{x\alpha +\beta \tan x}{x^2\tan x}=\frac{1}{3}\Rightarrow \underset{x\to 0}{lim}\frac{\alpha x+\beta (x+\frac{x^3}{3}+\cdots \infty )}{x^3\left(\frac{\tan x}{x}\right)}=\frac{1}{3}\Rightarrow \underset{x\to 0}{lim}\frac{(\alpha +\beta )x+\left(\frac{\beta }{3}\right)x^3+\cdots \infty }{x^3}=\frac{1}{3}$
So, $\alpha +\beta =0$
Also, $\beta =1\Rightarrow \alpha =-1$
Hence, ${\alpha }^2+{\beta }^2=2$