$$\frac{\sin \left(x\right)}{x}$$ Hello, I know how to prove without the definition, I tried a lot doing it by the definition of uniform continuity $\left|\frac{\sin \:\left(x\right)}{x}-\frac{\sin \:\left(y\right)}{y}\right|<\xi $ for every $x,y$ close enough. Thank you very much.
$\frac{\sin x}{x}$ uniform continuity by definition
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real-analysis
continuity
uniform-continuity
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0If $f(x)$ is a continuous function over $\mathbb{R}$ and $\lim_{x\to \pm\infty}f(x)=0$ then $f$ is U.C. over $\mathbb{R}$: it is a well-known lemma, a minor variation on "every continuous function over a compact interval is uniformly continuous". – 2017-01-29
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0Thanks that how i proved it, but im interesting if there is a way doing it directly from definition. – 2017-01-29
2 Answers
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Hint: Write $\frac{\sin x}{x} = \int_0^1 \cos (xt)\; dt$ and use a trigonometric identity on $\cos(xt)-\cos(yt)$ (as well as $|\sin(u)|\leq |u|$).
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We may prove that $f$ is Lipschitz-continuous. If $\left|x\right|\leq 1$ we have:
$$ \left|\frac{d}{dx}\frac{\sin x}{x}\right|\leq \sum_{n\geq 0}\frac{ |x|^{2n}}{(2n)!}=\cosh(|x|)\leq\cosh 1 $$ and if $|x|>1$ we have $$ \left|\frac{x\cos x-\sin x}{x^2}\right|\leq \frac{\sqrt{x^2+1}}{x^2}\leq\sqrt{2}$$ by the Cauchy-Schwarz inequality.