Can you help me to find the limits? $$\lim_{x \to 0^{+}}\frac{x}{a}\left[\frac{b}{x}\right] , \quad \lim_{x \to 0^{+}}\frac{b}{x}\left[\frac{x}{a}\right], \quad a,b>0$$ And what happens when $x \to 0^{-} $ ?
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1$\begingroup$ What does $[\frac{b}{x}]$ mean, is that a special notation for something? $\endgroup$– MartingaloFeb 10, 2014 at 22:48
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1$\begingroup$ I guess it's the biggest integer $\endgroup$– mjb4Feb 10, 2014 at 22:49
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$\begingroup$ It is the integral part!!! $\endgroup$– evindaFeb 10, 2014 at 22:50
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2 Answers
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For the first: we have $$\Bigl[\frac{b}{x}\Bigr]=\frac{b}{x}-\varepsilon(x)\ ,$$ where $0\le\varepsilon(x)<1$. Hence $$\frac{x}{a}\Bigl[\frac{b}{x}\Bigr]=\frac{b}{a}-\frac{x}{a}\varepsilon(x) \to\frac{b}{a}$$ as $x\to0^+$, beacuse the second term lies between $0$ and $x/a$.
For the second, as soon as $x<a$ we have $[x/a]=0$ and so $$\frac{b}{x}\Bigl[\frac{x}{a}\Bigr]=0\to0$$ as $x\to0^+$.
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$\begingroup$ Since we know that $[x]\leq x \geq [x]+1$,could we use this relation instead of the one that you used?? I applied it and got: $$\frac{b}{a}-\frac{x}{a} \leq \frac{x}{a}[\frac{b}{x}] \geq \frac{b}{a} $$..Can we say that the first limit is $0^{+}$ ..And,what's about $x\to O^{-}$.Is this the same result?? $\endgroup$– evindaFeb 10, 2014 at 23:07
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$\begingroup$ The second $\ge$ in your comment should be $<$. If you fix this it should be OK. I haven't looked at the $x\to0^-$ case. The same ideas should work, though the answers could possibly be different because for negative $x$ the integer part rounds away from $0$. For example $[\frac{3}{2}]$ is $1$ but $[-\frac{3}{2}]$ is $-2$, not $-1$. See what you can do. $\endgroup$– DavidFeb 10, 2014 at 23:21
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$\begingroup$ Using the same relation for the second limit,I have found the following: $$\frac{b}{a}-\frac{b}{x}\leq \frac{b}{x}[\frac{x}{a}] \leq \frac{b}{a}$$ What does this mean??That the limit does not exist???Or have I done something wrong?? $\endgroup$– evindaFeb 10, 2014 at 23:35
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1$\begingroup$ If you are trying to use the Sandwich Theorem (Pinching Theorem) it will not apply here and you will have to use other methods. $\endgroup$– DavidFeb 10, 2014 at 23:57
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$\begingroup$ I got stuck right now...Could you give me a hint?? $\endgroup$– evindaFeb 11, 2014 at 0:01
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This Limit does not exists!
Just look at the subseries $x_n=n$ and $x'_n := 0.9+n$
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$\begingroup$ Which limit do you mean that does not exist???the second one?? $\endgroup$– evindaFeb 10, 2014 at 23:10
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$\begingroup$ Could you check what I have done? Since we know that $[x]\leq x \leq [x]+1$,could we use this relation instead of the one that you used?? I applied it and got: $$\frac{b}{a}-\frac{x}{a} \leq \frac{x}{a}[\frac{b}{x}] \geq \frac{b}{a} $$..Can we say that the first limit is $0^{+}$ ..And,what's about $x\to O^{-}$.Is this the same result?? $\endgroup$– evindaFeb 10, 2014 at 23:13
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$\begingroup$ ah sorry im totally wrong! I did not see $x\to 0$, they both exists. Dont make it so complex! wait i edit my post $\endgroup$– mjb4Feb 10, 2014 at 23:16