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Does the sequence of functions defined by $f_{n}(x)=(1+x^{2n})^{1/2n}$ converge uniformly on $\mathbb{R}$.

For testing uniform convergence i know if the sequence $x_{n} = \sup \: \{ |f_{n}(x)-f(x) | : x \in \mathbb{R}\}$ converges to $0$ then $f_{n} \to f$ uniformly. But I don't know how to actually apply this result.

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    You could (should?) first identify $f$.2012-05-24
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    In other words, you should first figure out whether or not $f_n$ converges pointwise -- meaning whether or not $\lim f_n(x)$ makes sense for every $x \in \mathbb{R}$. If it does, then set $f(x) = \lim f_n(x)$, and see if you can apply the result you stated (regarding $x_n$).2012-05-24
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    Also notice that if $x\neq 0$, $f_n(x) = |x| (1+x^{-2n})^{\frac{1}{2n}}$.2012-05-24
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    What is the source of the problem?2012-05-24
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    Agree with Didier. The fun begins with determining what $f$ is. Notice that (before you take the root) when $|x|<1$ the constant $1$ dominates, whereas when $|x|>1$ the term $x^{2n}$ dominates for large indices $n$.2012-05-24
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    There's also the estimate $(1+|\theta|)^{\frac{1}{2n}} \leq 1+ \frac{|\theta|}{2n}$.2012-05-24
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    Well, thanks to various commenters, all the pieces are now on the table, so to speak, and what is left is to put them together. Let us hope this moment of true pedagogy is not ruined by the appearance of a full-bolts-on solution.2012-05-24
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    Now I'll never get my free T-shirt...2012-05-24
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    @Didier: I apologize :-)2012-05-24
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    @robjohn Apologies duly noted. However: Why did you post a full answer?2012-05-24
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    @Didier: oops! I didn't realize it was a homework problem (is it?).2012-05-24
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    @Josh: did you intend $\displaystyle(1+x^{2n})^{1/(2n)}$ (as everyone has been assuming)?2012-05-24
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    @robjohn Perhaps it's reasonable to undelete your answer after more than a year?2013-06-07
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    @Lord_Farin: thanks for the reminder. I lost track :-)2013-06-07
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    Try this [approach](http://math.stackexchange.com/questions/370023/how-to-prove-a-sequence-of-a-function-converges-uniformly/370071#370071).2013-06-07

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