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I want to show that $f_n(x)=\frac{x^2}{x^2+n}$ does not converge uniformly on $\mathbb{R}$, but it converges uniformly on any bounded subset $E\subseteq \mathbb{R}$.

First, note that $\displaystyle \lim_{n\rightarrow \infty} \frac{x^2}{x^2+n}=\lim_{n\rightarrow \infty} \frac{x^2/n}{x^2/n+1}=0$, so $f_n \rightarrow 0$ pointwise.

However, $f_n(\sqrt{n})=\frac{n}{2n}=\frac{1}{2},$ and thus, for any $\epsilon \in (0, 1/2)$, $\,$ $\displaystyle\sup_{x\in \mathbb{R}} |f_n(x)|\geq 1/2$, so $f_n$ cannot converge uniformly to $0$.

I am having trouble showing that $f_n$ converges uniformly on any bounded subset subset $E\subseteq \mathbb{R}$. If I take the derivative of $f_n$ I obtain

$f_n'(x)=\frac{2x(x^2+n)-x^2(2x)}{(x^2+n)^2}=\frac{2xn}{(x^2+n)^2}$. The derivative equals zero when $x=0$, but negative for any $x<0$ and positive for any $x>0$. This means that $f_n$ is decreasing for all $x<0$ and increasing for all $x>0$. How can this mean it converges uniformly to zero? Help! Thank you!!

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    $\begingroup$ You don't need to consider the derivative, try instead to bound $|f_n(x) - 0| = \frac{x^2}{x^2+n}$ independent of $x$. For example $\frac{1}{x^2+n} \leq \frac{1}{n}$ and $x^2 \leq C$ for some constant $C$ since $E$ is bounded. $\endgroup$ – Winther Jan 30 '17 at 2:08
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The idea is that taking $x$ large maximizes the functions. Indeed, note that $\lvert f_n(x) \rvert \le 1$ for all $x \in \mathbb R$ and all $n \in \mathbb N$. However, $$\lim_{x\to\pm\infty} f_n(x) = 1$$ so the maximum of $1$ is "achieved" for any $n$ when $x \to \pm \infty$. If we restrict $x$ to a bounded set, we preclude the possibility of letting $x\to\infty$. Indeed, if $E \subseteq \mathbb R$ is bounded, then there is an $L > 0$ such that $E \subseteq [-L,L]$. But, as you pointed out, $f_n$ is increasing as $\lvert x \rvert$ increases. Thus $$\lvert f(x) \rvert \le \frac{L^2}{L^2 + n}, \,\,\,\, \forall x \in E.$$ For fixed $\epsilon > 0$, we can take $N \in \mathbb N$ such that $\frac{L^2}{L^2 + N} < \epsilon$. Note, this $N$ does not depend on any individual $x$; only on $E$. Then for $n \ge N$, we have $$\lvert f_n(x) \rvert \le \epsilon \,\,\,\, \forall x \in E$$ wihch gives $$\| f_n \|_\infty < \epsilon$$ for $n \ge N$. This shows that $f_n$ converges uniformly to zero on $E$.

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