[tex] \frac{n}{4}-1 \leq [ \frac{n}{4} ] \leq \frac{xn}{4}
[/tex]
[tex] \lim_{n \to \infty}( \frac{n}{4}-1) \leq \lim_{n \to \infty} [ \frac{n}{4} ] \leq \lim_{n \to \infty} \frac{n}{4} [/tex] |:n (imparti fiecare chestie la n)
[tex] \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} [/tex]
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.
