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Prove that $\lim_{x\to \infty}\frac{3}{x}\lfloor\frac{x}{4}\rfloor=\frac{3}{4}$


If i put $x\to\infty$,the $\frac{3}{x}$ tends to zero and the $\lfloor\frac{x}{4}\rfloor$ tends to $\infty$.I do not know how they multiplied to get finite $\frac{3}{4}$.

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  • $\begingroup$ The idea is that $\frac3x\frac x4=\frac34$. The real problem is to show that the floor function doesn't mess that up too much for large $x$. $\endgroup$ – Arthur Jan 22 '16 at 16:10
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Hint $$\frac{x}{4}-1 \leq \left\lfloor\frac{x}{4}\right\rfloor \leq \frac{x}{4}$$

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  • $\begingroup$ Do we need to apply Sandwich theorem of limits? $\frac{x-1}{x}\leq\frac{\lfloor x \rfloor}{x}\leq\frac{x}{x}$ that means $\frac{\lfloor x \rfloor}{x}$ is sandwiched to limit $1$ at infinity.Am i right?@N.S. $\endgroup$ – Brahmagupta Jan 22 '16 at 16:19
  • $\begingroup$ @Brahmagupta Yes. $\endgroup$ – N. S. Jan 22 '16 at 16:27
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write [x/4]=(x/4)-{x/4} now regroup the terms.From the first limit you get the answer. Coming to the second limit {x/4} always lies in the range (0,1) so it is zero. This completes the proof.

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