Let $f$ be a function defined by $$f(x)=x^2 \sum_{n=0}^\infty \frac{1}{(x^2+1)^n}.$$ Find $f(0)$ and $\lim_{x\to 0}f(x)$.
I think as the convergence set of the series is $\mathbb{R}-\{0\}$, the function is not defined at zero. In the convergence set, $f(x)=x^2+1$ and the desired limit is $1$.
Am I right? Please help if my answer to the question is wrong. Thank you.
Let $ x\in\left(0,1\right] $, we have :
$$ \left|f\left(x\right)-\sum_{k=0}^{\left\lfloor\frac{1}{x}\right\rfloor}{\frac{x^{2}}{\left(1+x^{2}\right)^{k}}}-1\right|=1-\sum_{k=\left\lfloor\frac{1}{x}\right\rfloor+1}^{+\infty}{\frac{x^{2}}{\left(1+x^{2}\right)^{k}}}=1-\left(1+x^{2}\right)^{-\left\lfloor\frac{1}{x}\right\rfloor}\underset{x\to 0^{+}}{\longrightarrow}0 $$
Thus : \begin{aligned}\lim_{x\to 0^{+}}{f\left(x\right)}&=\lim_{x\to 0^{+}}{\left(1+\sum_{k=0}^{\left\lfloor\frac{1}{x}\right\rfloor}{\frac{x^{2}}{\left(1+x^{2}\right)^{k}}}\right)}\\ &=\lim_{x\to 0^{+}}{\left(1+x^{2}+1-\left(1+x^{2}\right)^{-\left\lfloor\frac{1}{x}\right\rfloor}\right)}\\ &=1\end{aligned}