For the third step I don't understand how they have worked out $C_1^1=0$? where did they get the value of 0 from?
p11(n)=(c01+c11n)lamda 1^n. Using p11(1)=1-a and c01=1, n=1,lamda 1=1 and lamba 1^1 =1 and so in the p11(n) equation I get c11 to be -a and not 0.
Any help would be much appreciated.
The $p_{11}(0)=1$ comes from definition $$p_{11}(0) = \Pr\left(X_0 = 1 \mid X_0 =1\right) = 1$$ and so does the second equation $$p_{11}(1) = \Pr\left(X_1 = 1 \mid X_0 =1\right) = \mathbb{P}_{11} = 1-\alpha$$
You can get the third step by setting $\beta = -\alpha+\epsilon$ and considering the limit $\epsilon \to 0$