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For the limit $$\lim_{x \to 0} y^x = 1$$ at what values of $y$ does it uniformly converge?

I was able to work out that if $y$ is bounded $[a, 1]$ (where $a>0)$, then it uniformly converges, and that if it's bounded $[1, b]$ (where $b>1)$, then it also uniformly converges.

I'm trying to find out whether or not it uniformly converges if $y$ gets arbitrarily big or if $y$ gets arbitrarily small.

sedrick
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    For every fixed nonzero $x$, $\lim\limits_{y\to0}y^x\ne1$ and $\lim\limits_{y\to\infty}y^x\ne1$. This fact suffices to prove that the sets of uniform convergence are subsets of compact sets of $(0,+\infty)$, that is, each set where the convergence is uniform is included in some $[a,b]$ for some $0<a<b<\infty$. – Did Apr 29 '18 at 08:04

1 Answers1

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Uniform convergence on an interval $(0,a)$ would require, for any $\epsilon > 0$, the existence of $\delta > 0$ such that if $0 < x < \delta$ then $|y^x -1|= |e^{x \log y}-1| < \epsilon$ for all $y \in (0,a)$.

However, given $\epsilon = (1 - e^{-1})/2$ and any $x \in (0, \delta)$, we can choose $y_x = e^{-1/x} \in (0,e^{-1/\delta})$ such that $|e^{x \log y_x}-1| = 1 - e^{-1} > \epsilon$. Hence, convergence cannot be uniform on an interval with $0$ as an endpoint.

See if you can make a similar argument to disprove uniform convergence for $y \in (b, \infty)$.

RRL
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