Consider the sequence of real-valued functions $\{f_n\}$ defined by $$f_n(x)=\frac{1}{1+nx^2}. $$ Assuming the fact that $\{f_n\}$ converges uniformly to a function find out all real numbers $x $ for which $$\displaystyle f'(x)=\lim_{n\rightarrow \infty}f'_n(x). $$
I calculated that $f_n(x)\rightarrow \begin{cases}0, \quad x\neq 0\\ 1, \quad x=0\end{cases}=f(x)$. We know that sequence of continuous functions which converge uniformly, must converge to a continuous function. But here $f$ is not continuous in the first place so how can we proceed ?
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Using the fact that $f_n\rightarrow f$ uniformly for $x\neq 0$, we have $\displaystyle f'_n(x)=\frac{-2xn}{(1+nx^2)^2}$ and $f'(x)=0 \,\forall x$. So $$\displaystyle\lim_{n\rightarrow\infty}f'_n(x)=\lim_{n\rightarrow\infty}\frac{-2xn}{(1+nx^2)^2}=\lim_{n\rightarrow\infty}\frac{\frac{-2x}{n}}{(\frac{1}{n}+x^2)^2}\rightarrow 0$$ irrespective of $x$. Thus the answer for this would be $\forall x\neq0$