3.
[tex]A(1)=\left(\begin{array}{ccc}1&1&1\\1&1&1\\1&1&1\end{array}\right)\\\\A(2)=\left(\begin{array}{ccc}1&2&2\\2&1&2\\2&2&1\end{array}\right)\\\\A(1)\times A(2)=\left(\begin{array}{ccc}1&1&1\\1&1&1\\1&1&1\end{array}\right)\times \left(\begin{array}{ccc}1&2&2\\2&1&2\\2&2&1\end{array}\right)=\left(\begin{array}{ccc}5&5&5\\5&5&5\\5&5&5\end{array}\right)=5\times A(1)[/tex]
4.
[tex]A\times B=\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)\times \left(\begin{array}{ccc}0&0&1\\0&1&0\\1&0&0\end{array}\right)=\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)[/tex]
[tex]B\times A=\left(\begin{array}{ccc}0&0&1\\0&1&0\\1&0&0\end{array}\right)\times \left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)=\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)[/tex]
A×B=B×A
5.
[tex]f(x)=\frac{x^2-3}{x^2+3}[/tex]
[tex]\lim_{x\to \infty} \frac{x^2-3}{x^2+3} =\frac{\infty}{\infty} \ aplicam \ L'Hospital[/tex]
Derivam sus, derivam jos si obtinem:
[tex]\lim_{x\to \infty} \frac{(x^2-3)'}{(x^2+3)'} =\lim_{x\to \infty} \frac{2x}{2x} =1[/tex]
