 Solution

For distinct eigenvalues, the formula holds for $n=0,1$. If the lower-left entry of $L^n$ is $b_n=({\lambda }_1^n-{\lambda }_2^n)/({\lambda }_1-{\lambda }_2)$, multiplication by $L$ gives $b_{n+1}={\lambda }_1{b}_n+{\lambda }_2^n=({\lambda }_1^{n+1}-{\lambda }_2^{n+1})/({\lambda }_1-{\lambda }_2)$. This proves by mathematical induction thatSumming the absolutely convergent matrix exponential series givesFor the repeated eigenvalue ${\lambda }_1={\lambda }_2=\lambda$, the continuous limit is $A(t)=e^{\lambda t}\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$. This is also the exponential of a two-dimensional Jordan block.