Does the sequence$ f_n(x)=\frac{x}{1+nx^2}$ converge uniformly on $\mathbb{R}$?

Question

I have to figure out if the sequence $f_n(x)=\dfrac{x}{1+nx^2}$ converges uniformly on $\mathbb{R}$

You should begin by deciding whether or not the function converges pointwise (and if so, what does it converge to)? — Ben Grossmann, Dec 01 '14 at 04:16
Get a candidate (pointwise) limit function $f(x)$. Then remember that $f_n(x)\to f(x)$ uniformly on $\mathbb R$ iff $\sup_{x\in\mathbb R}|f_n(x)-f(x)|\to 0$ as $n\to \infty$. — JohnD, Dec 01 '14 at 04:18

score 2 · Accepted Answer · edited Jan 07 '17 at 01:22

If $x = 0$, then $f_n(0) = 0, \forall n\in\mathbb{N}$. If $x \neq 0$, then by the inequality of arithmetic and geometric series, $$|f_n(x)| \leq \dfrac{|x|}{2|x|\sqrt{n}} = \dfrac{1}{2\sqrt{n}} \to 0 \text{ as } n \to \infty.$$ Thus, we take $f(x) = 0$ as the pointwise limit. We have $$f_n'(x) = \dfrac{1-nx^2}{(1+nx^2)^2} = 0 \iff x = \dfrac{1}{\sqrt{n}}.$$ Hence $$\max{f_n(x)} = f_n\left(\frac{1}{\sqrt{n}}\right) = \dfrac{1}{2\sqrt{n}}.$$ Lastly $$\sup{|f_n(x)-0|} = \dfrac{1}{2\sqrt{n}} \to 0 \text{ as } n\rightarrow\infty,$$ which, by an important theorem, proves uniform convergence on $\mathbb{R}$.

Does the sequence$ f_n(x)=\frac{x}{1+nx^2}$ converge uniformly on $\mathbb{R}$?

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