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Q. 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 $f$ find out real numbers $x$ for which $$f'(x)=\lim\limits_{n \to \infty} f'_n(x).$$

My answer is : $x \in R$,then $f'(x)=\lim\limits_{n \to \infty} f'_n(x)$

Is it correct? Please give me hints/solution.

Thanks in advance.

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    $\begingroup$ Wouldn't $f$ have to be continuous? $\endgroup$ – Hagen von Eitzen Apr 29 '18 at 20:55
  • $\begingroup$ @HagenvonEitzen ya f will not continious on$ x= 0$ $\endgroup$ – user396850 Apr 29 '18 at 20:57
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    $\begingroup$ which means that the convergence cannot be uniform (or the domain in question is not all of $\Bbb R$) $\endgroup$ – Hagen von Eitzen Apr 29 '18 at 20:59
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You know that $(f_n)$ converges to a function. Can you see what the function is by letting $n\rightarrow \infty$? From there can you find $f'$? Then do the same for $(f_n')$ to get an idea of how this sequence behaves.

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    $\begingroup$ There is no content in this answer. $\endgroup$ – Rene Schipperus Apr 29 '18 at 20:56
  • $\begingroup$ Sometimes showing how to get an answer is better than giving the answer. $\endgroup$ – marty cohen Apr 29 '18 at 20:59
  • $\begingroup$ I don't see your point: The idea is to get what $f(x)$ is like - from then you deduce that it is discontinuous at $0$, hence this is your problematic point. You can find $f'_n(x)$ and you can similarly show that you get pointwise convergence to $0$ which is $f'(0)$. Those were the $4$ hints I gave and they are all you need. I thought they were sufficient $\endgroup$ – asdf Apr 29 '18 at 23:04
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$$f(x) = \begin{cases} 0, & \text{if $x \neq 0$} \\ 1, & \text{if $x=0$} \end{cases}$$ is discontinuous.

so, $x$ must be belong $(-\infty,-a) \cup (a,+\infty)$ where $a \neq 0$ and $a>0$.

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