$\lim_{x\to 0} {[\sqrt{2} x]\over x}=\sqrt{2}-\lim_{x\to 0}{\{\sqrt{2} x\}\over x}$
Now $\lim_{x\to 0}{\{\sqrt{2} x\}\over x}=0$, how will I make understand to my school students who do not know the epsilon delta definition of limit.
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you can use sequence $a_n\rightarrow 0$. – Bobby Feb 22 '14 at 05:48
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Do the brackets $[\dots]$ and braces ${\dots}$ represent anything special? Also, is the radical meant to also cover the $x$? Thanks! – Christian Chapman Feb 22 '14 at 05:48
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that is box function i mean greattest integer function, and ${x}$ denote fractional part of $x$ – Myshkin Feb 22 '14 at 12:39
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If $[x]$ is the floor function and $\{x\} = x - [x]$ then I think you are wrong. In fact $[\sqrt 2 x]=0$ if $x \to 0^+$ while $[\sqrt 2 x] = -1$ if $x\to 0^-$ hence the given limit does not exist.
Emanuele Paolini
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