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 August 31st, 2009, 12:25 PM #1 Newbie   Joined: Jul 2009 Posts: 11 Thanks: 0 limit Show that the function$f(x)= x^2 \sin \left( {\frac{1}{{x^2 }}} \right)\$ is bordered! bordered mean :example:$f(n) = \frac{n}{{n + 1}},n \in N \Rightarrow \frac{n}{{n + 1}} < 1 \$ I dont know the typical word
 August 31st, 2009, 01:03 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: limit The word you are looking for is bounded. $f(x)$ is not defined at $x=0.$ For $0<|x|\leq1,$ we have $\left|\sin\left(\frac1{x^2}\right)\right|\leq1$ and $x^2\leq1,$ so $|f(x)|\leq1.$ For $|x|>1,$ we know that $\sin\left(\frac1{x^2}\right)<\frac1{x^2},$ so $|f(x)|<1.$ Therefore $-1 for all $x\in\mathbb{R}\setminus\{0\}.$ But it is possible to find better bounds for this function - I'll leave that to you or someone else if you want to find them.

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