proof: the sequence $\{x_n\}_{n=1}^{\infty}$ converges, and find $\lim x_n$. (check).
Question: Suppose $x_1= \frac 12$ and $x_{n+1}=x_n^2$. Show that $\{x_n\}$ converges and find $\lim x_n$. Hint: you cannot divide by zero!
From the question, I know that $\{x_n\}=\{\frac 12, \frac 14, \frac 1{16}, ....\} = \{\frac 1{2^{2^n}}: n\in N\}$.
Then, $\lim x_n = 0$. So, to test convergence, let $\varepsilon > \frac 1{2^{2^M}}$, and $M \in N$. Then, for $\varepsilon >0$, there exists $M$ such that$|\frac 1{2^{2^n}}-0|<\varepsilon $ for $n\ge M$ because $2^{2^M}\le2^{2^n} \rightarrow \frac 1{2^{2^M}}\ge \frac 1{2^{2^n}} \rightarrow \frac 1{2^{2^n}}<\varepsilon$.
Could you check this proof is fine? I am not sure about my answer since I don't know when I make use of hint.
Thank you in advance.
I edited. Thanks for pointing out. Could you check anything wrong??