Limit exercise mixed with linear algebra asked by Dan Sitrua in many math groups

Exercise:

Let \(A,B\in {{M}_{2}}\left( \mathbb{R} \right)\) such that ${{A}^{2}}-5A={{B}^{2}}-5B=-6I$

Find $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\det \left( {{A}^{n}}-{{B}^{n}} \right)}{{{9}^{n}}}$

Solution: we will use induction to find factorization for ${{A}^{n}}-{{B}^{n}}$

We have ${{A}^{2}}-5A={{B}^{2}}-5B\Leftrightarrow {{A}^{2}}-{{B}^{2}}=5\left( A-B \right)=\left( {{3}^{2}}-{{2}^{2}} \right)\left( A-B \right)$

Now \({{A}^{2}}-5A=-6I\overset{\times A}{\mathop{\Leftrightarrow }}\,{{A}^{3}}-5{{A}^{2}}=-6A\) and ${{B}^{3}}-5{{B}^{2}}=-6B$

So  ${{A}^{3}}-{{B}^{3}}=5{{A}^{2}}-5{{B}^{2}}-6A+6B\Leftrightarrow {{A}^{3}}-{{B}^{3}}=5\left( {{A}^{2}}-{{B}^{2}} \right)-6\left( A-B \right)$

$\Leftrightarrow {{A}^{3}}-{{B}^{3}}=25\left( A-B \right)-6\left( A-B \right)=19\left( A-B \right)=\left( {{3}^{3}}-{{2}^{3}} \right)\left( A-B \right)$

So we deduce that , ${{A}^{n}}-{{B}^{n}}=\left( {{3}^{n}}-{{2}^{n}} \right)\left( A-B \right)$ , $n\in {{\mathbb{N}}^{\ge 2}}$

Thus $\det \left( {{A}^{n}}-{{B}^{n}} \right)=\left( {{3}^{n}}-{{2}^{n}} \right)\det \left( A-B \right)$

Hence $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\det \left( {{A}^{n}}-{{B}^{n}} \right)}{{{9}^{n}}}=\det \left( A-B \right)\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{3}^{n}}-{{2}^{n}}}{{{3}^{2n}}}=0$ why ??

Note that, $\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{3}^{n}}-{{2}^{n}}}{{{3}^{2n}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{3}^{n}}}{{{3}^{2n}}}-\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{2}^{n}}}{{{3}^{2n}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{2}^{n}}}{{{3}^{2n}}}=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \frac{2}{9} \right)}^{n}}$

Let $w={{\left( 2/9 \right)}^{n}}\Leftrightarrow \ln \left( w \right)=n\ln \left( 2/9 \right)=-n\ln \left( 9/2 \right)$

So $w=\underset{n\to \infty }{\mathop{\lim }}\,{{e}^{-n\ln \left( 9/2 \right)}}={{e}^{-\infty }}=0$ as Exponential is continuous function

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the idea of solution Credit to Vulvaal Pal