Uniform convergence and derivatives

Hint

It's pretty clear that $(f_n)$ is pointwise convergent to the zero function $f$ on $\mathbb{R}$ and we have $$f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}=0\iff x=\frac{1}{\sqrt{n}}=x_n$$ and since $$||f_n-f||_\infty=|f_n(x_n)|=\frac{1}{2x_n}\to0$$ so $(f_n)$ is uniformly convergent to $f$ on $\mathbb{R}$

Finally we can see easily that for $x=0$, $f'_n(0)=1\to 1\neq f'(0)=0$ and for $x\neq 0$ $f'_n(x)\to 0=f'(x)$

Let's prove that $f_n$ converges uniformly to 0 in $\mathbb R$.

From the definition:

$$\sup\limits_{x\in\mathbb R}|f_n(x)-0|=\sup\limits_{x\in\mathbb R}\dfrac{|x|}{|1+nx^2|}=$$ $$\sup\limits_{x\in\mathbb R}\dfrac{|x|}{1+\sqrt{n}|x|^2}\leq\sup\limits_{x\in\mathbb R}\dfrac{|x|}{2\sqrt{n}|x|}=\dfrac{1}{2\sqrt{n}}\overset{n\rightarrow\infty}{\longrightarrow }0$$ Therefore, $\sup\limits_{x\in\mathbb R}|f_n(x)-0|\rightarrow 0, $ i.e. $f_n$ converges uniformly to $0$.

Now, take $f_n'(x)=\dfrac{1-nx^2}{(1+nx^2)^2}$

• If $x=0$ then $f_n'(0)=1\nrightarrow f'(0)=0$
• If $x\neq 0$ then you can easily check that $f_n'(x)\rightarrow f'(x)=0$

Since you have to calculate derivatives anyway, you might as well apply the standard method from calculus (argue about the sign of $f'_n$ on three intervals) to show that $\max (|f_n(x)| : x\in \mathbb{R})=\dfrac{1}{2\sqrt{n}}$ and argue from there about uniform convergence. Obviously, you want to show that $f'_n(x)\to 0$ if $x\neq 0$ and $f'_n(0)$ does not. The nice thing about $f_n'(0)$ is that it's a constant sequence.