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.
Proving the uniform convergence of $\sin(\frac{x}{n}) e^{-x^2}$
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real-analysis
analysis
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0It'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. :-) – 2011-09-15
1 Answers
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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.