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How might one show that $\left(\sin(\frac{x}{n})e^{-x^2}\right)_n$ converges uniformly? I tried finding the supremum by setting the 1st derivative to $0$, but that gives a hard-to-solve equation. There must therefore be a looser bound, but I am not seeing it. Thanks. 

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HINT Use the fact that $|\sin (\frac{x}{n})| \leq |\frac{x}{n}|$. – Srivatsan Sep 15 '11 at 17:12
Thanks, Srivatsan! How did you spot that? :-) – jose Sep 15 '11 at 17:22
It's a standard idea, worth remembering. As a secondary point, I am a fan of avoiding differentiation as far as possible, so I like these approximations. :-) – Srivatsan Sep 15 '11 at 17:24
up vote 6 down vote accepted

Notice that $xe^{-x^2}$ is uniformly bounded, say by some constant $M$. Then $$\biggr|\sin\left(\frac{x}{n}\right)e^{-x^2}\biggr|\leq \biggr|\frac{x}{n}e^{-x^2}\biggr|\leq \frac{M}{n}.$$

As $\frac{1}{n}$ converges and this no longer depends on $x$ we have uniform convergence.

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