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

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.

• Wouldn't $f$ have to be continuous? – Hagen von Eitzen Apr 29 '18 at 20:55
• @HagenvonEitzen ya f will not continious on$x= 0$ – user396850 Apr 29 '18 at 20:57
• which means that the convergence cannot be uniform (or the domain in question is not all of $\Bbb R$) – Hagen von Eitzen Apr 29 '18 at 20:59

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.
• 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 – asdf Apr 29 '18 at 23:04
$$f(x) = \begin{cases} 0, & \text{if x \neq 0} \\ 1, & \text{if x=0} \end{cases}$$ is discontinuous.
so, $x$ must be belong $(-\infty,-a) \cup (a,+\infty)$ where $a \neq 0$ and $a>0$.