$A = \left[\begin{matrix}\cos \alpha & \sin \alpha \\- \sin \alpha & \cos \alpha \\ \end {matrix} \right]$
$A^n = \left[\begin{matrix}\cos n \alpha & \sin n \alpha \\- \sin n \alpha & \cos n \alpha \\ \end {matrix}\right]$
$\frac{1}{n}A^n = \left[\begin{matrix}\frac{\cos n \alpha}{n} & \frac{\sin n \alpha}{n} \\- \frac{\sin n \alpha}{n} & \frac{\cos n \alpha}{n} \\ \end{matrix}\right]$
$\lim_{n \rightarrow \infty}\frac{1}{n}A^n$
$\lim_{n \rightarrow \infty} \left[\begin{matrix}\frac{\cos n \alpha}{n} & \frac{\sin n \alpha}{n} \\- \frac{\sin n \alpha}{n} & \frac{\cos n \alpha}{n} \\ \end{matrix}\right]$
$\lim_{n \rightarrow \infty} \frac{\sin n \alpha}{n} = 0$
$\lim_{n \rightarrow \infty} \frac{\cos n \alpha}{n} = 0$
$ = \left[\begin{matrix}0 & 0 \\0 & 0 \\ \end{matrix}\right]$
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$(P)$ જો $A \neq I_{2},$ હોય તો $|A|=-1$:
$(Q)$ જો $|\mathrm{A}|=1,$ હોય તો $\operatorname{tr}(\mathrm{A})=2$
જ્યાં $I_{2}$ એ $2 \times 2$ નો એકમ શ્રેણિક અને $\operatorname{tr}(A)$ એ શ્રેણિક $A$ ના અગ્ર વિકર્ણના ઘટકોનો સરવાળો દર્શાવે તો
$f(x) = {\rm{ }}\left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{e^x} - 1} \right)^2}}{{\sin {\mkern 1mu} \left( {\frac{x}{k}} \right){\mkern 1mu} \log {\mkern 1mu} \left( {1 + \frac{x}{4}} \right)}}{\mkern 1mu} ,{\mkern 1mu} x \ne 0}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 12{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ,x{\mkern 1mu} {\mkern 1mu} = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }
\end{array}} \right.$ એ સતત વિધેય હોય તો $k$ ની કિમંત મેળવો.