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2014-07-09T11:13:11+03:00
 \frac{n}{4}-1 \leq [ \frac{n}{4} ]  \leq  \frac{xn}{4}                                                                                 

 \lim_{n \to \infty}(  \frac{n}{4}-1) \leq  \lim_{n \to \infty} [ \frac{n}{4} ]  \leq  \lim_{n \to \infty}  \frac{n}{4}       |:n  (imparti fiecare chestie la n)
 \lim_{n \to \infty}  \frac{n-4}{4n}  \leq  \lim_{n \to \infty}  [\frac{n}{4}]* \frac{1}{n}   \leq  \lim_{n \to \infty}  \frac{n}{4n}
In cazul primei limite, observi ca ai infinit/infinit, asa ca aplici L'Hospital si obtii 1/4.
In cazul ultimei limite, se reduce n cu n si ramane 1/4.
Daca prima si ultima tind la 1/4, inseamna ca si [n/4]/n tinde tot la 1/4.