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Prove that $f_n(x)=(\sqrt{x^2+\frac{1}{n}})_{n\in\mathbb{N}}$ $(x\in\mathbb{R})$ is pointwise convergent, and then check to see if its uniformly convergent

So I can prove it is pointwise:

$f_n(x)=\sqrt{x^2+\frac{1}{n}} \rightarrow x$ as $n\rightarrow\infty$

Thus to prove uniform I need $|f_n(x_n)-f(x_n)|\rightarrow 0$ so I have $\left|\sqrt{x_n^2+\frac{1}{n}}-x_n\right|$ but I am unsure how to simplify this, and to what I should make $x_n$ equal to (i.e. would making it be $n$ simpler, as uniform convergence allows us to pick any $x_n$).

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  • $\begingroup$ The problem asks about $x\in \mathbb{R}$. If $x=-17$, the limit is $34$. $\endgroup$ – André Nicolas Jan 9 '16 at 18:12
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First note that, as Andre Nicolas hinted, $f_n(x) \to |x|$ since $\sqrt{x^2} = |x|.$

Next note that for any sequence $x_n$, $$ 0 \leq \sqrt{x_n^2 + \frac{1}{n}} - |x_n| = \frac{1}{n} \times \frac{1}{\sqrt{x_n^2+\frac{1}{n}}+|x_n|} = \frac{1}{\sqrt{n^2 x_n^2 + n} + |nx_n|} \leq \frac{1}{\sqrt{n}}. $$

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