Question

Calculați limita de mai jos:

Answer 1

[tex]\text{Vom aplica criteriul radicalului.}\\ \text{Notam }x_n=\dfrac{(n^2+1)(n^2+2^2)\ldots(n^2+n^2)}{n^{2n}}\\ \\ \text{Atunci: }\displaystyle\limit\lim_{n\to\infty} \sqrt[n]{x_n}=\displaystyle\limit\lim_{n\to\infty}\dfrac{x_{n+1}}{x_n}\\ \displaystyle\limit\lim_{n\to\infty}\dfrac{x_{n+1}}{x_n}=\displaystyle\limit\lim_{n\to\infty}\dfrac{[(n+1)^2+1][(n+1)^2+2^2]\ldots [(n+1)^2+(n+1)^2]}{(n+1)^{2n+2}}\\ \cdot \dfrac{n^{2n}}{(n^2+1)(n^2+2^2)\ldots(n^2+n^2)}\\ \text{Mai departe facem niste artificii de calcul:}[/tex]
[tex]\displaystyle\limit\lim_{n\to \infty} \dfrac{(n+1)^2+1}{(n+1)^2}\cdot \dfrac{(n+1)^2+2^2}{(n+1)^2}\cdot \ldots \cdot \dfrac{2\cdot (n+1)^2}{(n+1)^2}\cdot \dfrac{n^2}{n^2+1}\cdot \dfrac{n^2}{n^2+2^2}\cdot\\ \cdot \ldots \cdot \dfrac{n^2}{2\cdot n^2}=1\cdot 1\cdot 1\cdot\ldots\cdot 2\cdot 1\cdot 1\cdot \ldots \cdot \dfrac{1}{2}= \boxed{1}\\ \text{Deci limita ceruta este 1.} [/tex]