Question

Va rog mult sa ma ajutati!!!​

Answer 1

3.

$A(1)=\left(\begin{array}{ccc}1&1&1\\1&1&1\\1&1&1\end{array}\right)\\\\A(2)=\left(\begin{array}{ccc}1&2&2\\2&1&2\\2&2&1\end{array}\right)\\\\A(1)\times A(2)=\left(\begin{array}{ccc}1&1&1\\1&1&1\\1&1&1\end{array}\right)\times \left(\begin{array}{ccc}1&2&2\\2&1&2\\2&2&1\end{array}\right)=\left(\begin{array}{ccc}5&5&5\\5&5&5\\5&5&5\end{array}\right)=5\times A(1)$

4.

$A\times B=\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)\times \left(\begin{array}{ccc}0&0&1\\0&1&0\\1&0&0\end{array}\right)=\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)$

$B\times A=\left(\begin{array}{ccc}0&0&1\\0&1&0\\1&0&0\end{array}\right)\times \left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)=\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right)$

A×B=B×A

5.

$f(x)=\frac{x^2-3}{x^2+3}$

$\lim_{x\to \infty} \frac{x^2-3}{x^2+3} =\frac{\infty}{\infty} \ aplicam \ L'Hospital$

Derivam sus, derivam jos si obtinem:

$\lim_{x\to \infty} \frac{(x^2-3)'}{(x^2+3)'} =\lim_{x\to \infty} \frac{2x}{2x} =1$