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I'm trying to show that $f_n(x) = (x + \frac{1}{n})^2$ converges uniformly. So $f(x) = \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} (x + \frac{1}{n})^2 = (\lim_{n \to \infty} (x + \frac{1}{n}))^2 = x^2$.

I want to show that $|f_n(x) - f(x)| < \epsilon$ for $n$ large enough. So

$|f_n(x) - f(x)| = |(x + \frac{1}{n})^2 - x^2| = |x^2 + \frac{2x}{n} + \frac{1}{n^2} - x^2| = |\frac{2xn}{n^2} + \frac{1}{n^2}|$

Edit Uniform convergence does not occur on $\mathbb{R}$. But let me consider the interval $[-5, 5]$. Since $x \in [-5, 5]$, $|\frac{2xn}{n^2} + \frac{1}{n^2}| < |\frac{10n}{n^2} + \frac{1}{n^2}|$, which will be made less than $\epsilon$ for $n$ large enough.

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    Yes, sorry. This is the classical example, indeed. You are right Santiago. I was thinking on another thing. Sorry mainly for @Student. I erased that. Thanks a lot for pointing that. Silly me, hehe2012-05-01

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I think this sequence doesn't converge uniformly to $f(x)=x^2$. We need to prove that: given $\epsilon>0$, there exists $n_\epsilon$ such that $n>n_\epsilon$ implies $|f_n(x)-f(x)|<\epsilon$ for all $x\in\mathbb{R}$. But

$\left|\frac{2xn}{n^2} + \frac{1}{n^2}\right|=\left|\frac{2xn-1}{n^2}\right|$

and, for a fized $n$, this is less than $\epsilon$ for all $x$ if and only if

$|2xn-1|\leqslant n^2\epsilon$

wich is false, because $x$ can be large enough to make the left side greater than the right.

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    @matgaio: Awesome, thank you!2012-05-01
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consider $g_n(x)=x/n$ on whole $\mathbb{R}$, then $lim_{n\rightarrow\infty}x/n=0$, now fix $\epsilon>0$ $|x/n-0|=|x/n|<\epsilon$ whenever $n>x/\epsilon$ $\Rightarrow$ $n$ depends on both $x$ and $\epsilon$ so $g_n(x)$ is not Uniformly Convergent Over $\mathbb{R}$ by the definition of Uniform Convergence. I hope from This Hint you can do your problem.

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    I have edited my answer.thank you David.2012-05-01