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Find the limit by solving equation II


The idea is pretty much the same as Exercise 9.4.

Solution: Since $(t_n)$ converges. Set $\lim t_n=s$. By $t_{n+1}=\dfrac{t_n^2+2}{2t_n}$, we have
\begin{equation}\label{09-05-01}
\lim 2t_{n+1}t_n=\lim (t_n^2+2).
\end{equation} On one hand, by Theorems 9.2 and 9.4, we have
\begin{equation}\label{09-05-02}
\lim 2t_{n+1}t_n=2\lim t_{n+1}\lim t_n=2t^2.
\end{equation} On the other hand, by Theorems 9.3 and 9.4, we have
\begin{equation}\label{09-05-03}
\lim (t_n^2+2)=\lim t_n\lim t_n+\lim 2=t^2+2.
\end{equation} Combining \eqref{09-05-01}, \eqref{09-05-02}, \eqref{09-05-03}, we have
$$
2t^2=t^2+2.
$$ Solving for $t$, we must have $t=\pm \sqrt 2$.

Since $t_1>0$, it is clear by induction that $t_n>0$ for all $n$. Now it follows from Exercise 8.9 that $t>0$. Therefore, we conclude that $t=\sqrt 2$.


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